We can observe that result cannot exceed 201 — Awruk gets at least 101 votes from one person and Elodreip cannot get more than 100 votes from one person. So we can iterate over every possible integer from 1 to 201 and check if Awruk wins with k set to this integer. We have to remember that k — a_i is always at least 0, so we have to check this condition too. Complexity O(n * M), where M denotes maximum possible value of a_i. Try to solve it in O(n).

#include <bits/stdc++.h>

using namespace std;

int n;
int mx = 0, sum = 0;

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		int a;	scanf("%d", &a);
		mx = max(mx, a);
		sum += a;
	}
	
	sum *= 2;
	sum += n;
	sum /= n;
	
	printf("%d", max(sum, mx));
	return 0;
}

First, let's observe that we can replace array a_i with array b_i = a_i  -  a_i - 1, because all we care about are differences between neighboring elements. Now, we can see that our lost array can have length d if and only if for every j such that j  +  d  ≤  n, b_j  =  b_j + d. So we can iterate over every possible d from 1 to n and check if it is correct in O(n). Complexity of whole algorithm is O(n²).

#include <bits/stdc++.h>

using namespace std;

const int N = 1007;

int n;
int in[N];

bool ok(int d){
	for(int i = 0; i + d < n; ++i)
		if(in[i + 1] - in[i] != in[i + d + 1] - in[i + d])
			return false;
	return true;
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &in[i]);
	
	vector <int> res;
	for(int i = 1; i <= n; ++i)
		if(ok(i))
			res.push_back(i);
	
	printf("%d\n", res.size());
	for(int v: res)
		printf("%d ", v);
	return 0;
}

Basically in problem we are given a word in which for every i we can reverse prefix of first i elements and we want to get the smallest lexicographically word. We will show that we can always achieve word in form a^jb^n - j.

Let's say that we solved our problem for prefix of length i and for this prefix we have word a^jb^i - j (at the beginning it's just empty word). If our next letter is b then we do nothing, because we will get word a^jb^{i - j + 1} which is still the smallest lexicographically word. Otherwise we want to reverse prefix of length i, add letter a and reverse prefix of length i  +  1, so we get a word a^j + 1b^i - j, which is still fine for us.

There is still a problem — what if we have already reversed prefix i and we just said that we will reverse it second time. But instead of reversing it second time, we can deny it's first reverse.

Final complexity is O(n).

#include <bits/stdc++.h>

using namespace std;

const int N = 1007;

string s;
bool write[N];

int main(){
	cin >> s;
	for(int i = 1; i < s.size(); ++i)
		if(s[i] == 'a'){
			write[i - 1] ^= 1;
			write[i] = 1;
		}
	
	for(int i = 0; i < s.size(); ++i)
		printf("%d%c", write[i], i + 1 == (int)s.size() ? '\n' : ' ');
	return 0;
}

Deleting prefix and suffix is nothing more than taking a subarray. If subarray is common for all permutations then it has to appear in first permutation. We renumber all permutations such that first permutation is 1, 2, ..., n  -  1, n.

Now for every i in every permutation we count how long is subarray starting at i which looks like i, i  +  1, ..., i  +  k. It can be easily done in O(n) for one permutation with two pointers technique.

Now for every element i we compute reach[i] equal the longest subarray starting in i which looks like i, i  +  1, ..., i  +  k and it apears in all subarrays. It is just minimum over previously calculated values for all permutations.

Now we can see that our result is . Final complexity O(nm).

#include <bits/stdc++.h>

using namespace std;

typedef long long int LL;

const int N = 1e5 + 7;

int n, m;
int mn[N];
int ren[N];
int perm[15][N];

int main(){
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			scanf("%d", &perm[i][j]);
	
	for(int i = 1; i <= n; ++i)
		ren[perm[1][i]] = i;
	
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			perm[i][j] = ren[perm[i][j]];
		
	for(int i = 1; i <= n; ++i)
		mn[i] = n;
		
	for(int i = 1; i <= m; ++i){
		int cur = 1;
		for(int j = 1; j <= n; ++j){
			if(cur < j)
				++cur;

			while(cur < n && perm[i][cur + 1] == perm[i][cur] + 1)
				++cur;
			mn[perm[i][j]] = min(mn[perm[i][j]], perm[i][cur]);
		}
	}
	
	LL res = 0;
	int now = 1;
	while(now <= n){
		int cur = mn[now] - now + 1;
		res += 1LL * (cur + 1) * cur / 2LL;
		now = mn[now] + 1;
	}
	
	printf("%lld\n", res);
	return 0;
}

Let's compute result if there are no edges, we can add them later. If there are no edges then result for pair (i, j) is min(x_i  +  y_j, x_j  +  y_i). First let's fix i for which we want to compute result. Then calculate result with all pairs j such that x_i  +  y_j  ≤  x_j  +  y_i. After some transformations we get that x_i  -  y_i  ≤  x_j  -  y_j. Similarly we have that y_i  +  x_j  <  x_i  +  y_j if x_i  -  y_i  >  y_j  -  x_j.

So let's sort over differences of x_i  -  y_i and compute prefix sums of x_i and suffix sums of y_i. Now we can compute for every i result in O(1). Then we can iterate over every edge (u, v) and subtract min(x_u  +  y_v, x_v  +  y_u) from result of u and v.

Complexity O(nlogn).

#include <bits/stdc++.h>

using namespace std;

typedef long long int LL;

#define st first
#define nd second
#define PII pair <int, int>

const int N = 3e5 + 7;

int n, m;
PII diff[N];
int place[N];
vector <int> G[N];

LL ans[N];
int x[N], y[N];
LL pref[N], suf[N];

int main(){
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; ++i){
		scanf("%d %d", &x[i], &y[i]);
		diff[i] = {y[i] - x[i], i};
	}
	
	for(int i = 1; i <= m; ++i){
		int u, v;
		scanf("%d %d", &u, &v);

		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	sort(diff + 1, diff + n + 1);
	for(int i = 1; i <= n; ++i)
		place[diff[i].nd] = i;
	
	for(int i = 1; i <= n; ++i)
		pref[i] = pref[i - 1] + y[diff[i].nd];
	
	for(int i = n; i >= 1; --i)
		suf[i] = suf[i + 1] + x[diff[i].nd];
	
	for(int i = 1; i <= n; ++i){
		int u = diff[i].nd;
		LL res = pref[i - 1] + suf[i + 1] + 1LL * (i - 1) * x[u] + 1LL * (n - i) * y[u];
		
		for(int v: G[u])
			res -= min(x[u] + y[v], x[v] + y[u]);
		ans[u] = res;
	}
	
	for(int i = 1; i <= n; ++i)
		printf("%lld ", ans[i]);
	return 0;
}

First let's observe that if there exists valid subset then it's size is at most 7 (because product of 7 smallest primes is bigger then 3 * 10⁵). Let's define dp[i][j] — number of ways to pick i different elements such that their gcd is equal to j. We can use inclusion--exclusion principle to calculate it. Then dp[i][j] = — , where cnt_j denotes number of a_i such that j | a_i. Because for k * j  >  3 * 10⁵, dp[i][k * j]  =  0 we have to check only k * j  ≤  3 * 10⁵.

Our answer is the smallest i such that dp[i][1] is non-zero. Since dp[i][j] can be quite big we should compute it modulo some big prime.

Final complexity is O(logM * (n + M)), where M is equal to maximum of a_i.

#include <bits/stdc++.h>

using namespace std;

const int N = 3e5 + 7;
const int MX = 1e9 + 7;

int n;
int cnt[N];
int sil[N];
int odw[N];
int dp[20][N];

int fast(int a, int b){
	int ret = 1;
	while(b){
		if(b & 1)
			ret = (1LL * ret * a)%MX;
		
		b >>= 1;
		a = (1LL * a * a)%MX;
	}
	
	return ret;
}

int newton(int a, int b){
	if(b < 0 || a < b)	return 0;
	return (((1LL * sil[a] * odw[b])%MX) * odw[a - b])%MX;
}

void sub(int &a, int b){
	a -= b;
	if(a < 0)
		a += MX;
}

int main(){
	sil[0] = 1;
	for(int i = 1; i < N; ++i)
		sil[i] = (1LL * sil[i - 1] * i)%MX;
	
	odw[N - 1] = fast(sil[N - 1], MX - 2);
	for(int i = N - 1; i >= 1; --i)
		odw[i - 1] = (1LL * odw[i] * i)%MX;

	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		int a;
		scanf("%d", &a);
		cnt[a]++;
	}
	
	for(int i = 1; i < N; ++i)
		for(int j = i + i; j < N; j += i)
			cnt[i] += cnt[j];
	
	for(int i = 1; i < 20; ++i){
		for(int j = N - 1; j >= 1; --j){
			dp[i][j] = newton(cnt[j], i);
			for(int k = j + j; k < N; k += j)
				sub(dp[i][j], dp[i][k]);
		}
		
		if(dp[i][1] > 0){
			printf("%d\n", i);
			return 0;
		}
	}
	
	puts("-1");
	return 0;
}

#include <bits/stdc++.h>

using namespace std;

const int N = 3e5 + 7;

int n;
int cnt[N];
int roz[N];
int dist[N];
queue <int> Q;
vector <int> dv[N];

int base(int a){
	int ret = 1;
	while(a > 1){
		if(ret%roz[a] != 0)
			ret *= roz[a];
		a /= roz[a];
	}
	
	return ret;
}

void getEdges(int u, int d){
	vector <int> cur;
	vector <int> val;
	
	while(u > 1){
		cur.push_back(roz[u]);
		u /= roz[u];
	}
	
	for(int v: cur)
		u *= v;
	
	int T = 1 << (int)cur.size();
	val.resize(T);
	
	for(int i = 0; i < T; ++i){
		val[i] = u;
		for(int j = 0; j < (int)cur.size(); ++j)
			if(i & (1 << j))
				val[i] /= cur[j];
	}
	
	for(int i = 0; i < T; ++i){
		int s = 0;
		for(int j = i; true; j = (j - 1) & i){
			if(__builtin_popcount(i ^ j) & 1)
				s -= cnt[val[j]];
			else
				s += cnt[val[j]];
			
			if(j == 0)
				break;
		}
		
		assert(s >= 0);
		if(s && dist[val[i]] == -1){
			dist[val[i]] = d;
			Q.push(val[i]);
		}
	}
}

int main(){
	scanf("%d", &n);
	for(int i = 2; i < N; ++i){
		if(roz[i] != 0)
			continue;

		for(int j = i; j < N; j += i)
			roz[j] = i;
	}
	
	for(int i = 1; i < N; ++i)
		for(int j = i; j < N; j += i)
			dv[j].push_back(i);
	
	for(int i = 1; i < N; ++i)
		dist[i] = -1;
	
	for(int i = 1; i <= n; ++i){
		int a;
		scanf("%d", &a);
		a = base(a);
		
		if(dist[a] != -1)
			continue;

		dist[a] = 1;
		Q.push(a);
		
		for(int v: dv[a])
			cnt[v]++;
	}
	
	while(!Q.empty()){
		int u = Q.front();
		Q.pop();
		getEdges(u, dist[u] + 1);
	}
	
	printf("%d\n", dist[1]);
	return 0;
}

Let's solve the problem for some string s for any time.

Let's say, that partition of string s into k strings s_1s2... s_i1, s_i1 + 1... s_i2, ..., {s_{ik - 1 + 1}}... s_ik is good if at least one pair of this strings are equal. We want to find a minimal possible number of different strings in all good partitions.

It's easy to see, that the answer is  - 1 if and only if all symbols in s are different. And if we have two equal symbols s_i = s_j (i < j) we can cut a string into strings s₁... s_i - 1, s_i, s_i + 1... s_j - 1, s_j, s_j + 1... s_n and it is a good partition. In this partition there is at most 4 different strings.

So the answer can be  - 1, 1, 2, 3, 4.

The answer is  - 1 if all symbols in s are different (case 0).

The answer is 1 if the string s = aaa... a, for some string a (case 1).

The answer is 2 if the string s is aab, aba or baa for some strings a and b (case 2).

The answer is 3 if the string s is baac, bcaa or aabc for some strings a, b, c. In two last cases it's easy to see, that |a| = 1 (case 3).

To solve our problem let's build suffix array with lcp for string s. And let's find lt_i~--- minimal possible number r, such that s_isi + 1... s_r is a tandem (the string, that can be presented as aa for some string a) and rt_i~--- maximal possible number l such that s_lsl + 1... s_i is a tandem. This numbers can be found using Main and Lorentz algorithm for finding tandem repetitions in the string.

Now we can solve query for segment [l, r]: \begin{itemize} \item Case 0: if r - l ≥ 26, there exists equal symbols, otherwise we can check it by O(r - l); \item Case 1: to check that s[l... r] = aa... a we can see that |a| is a divisor of (r - l + 1) and (r - l + 1) / |a| is a prime number (if we take a longest possible string a). So we should check only O(log(n)) lenghts of string a; \item Case 2: s = aab lt_l ≤ r, s = baa rt_r ≥ l. In the last case we should check, that s[l... r] has a border. It's the most interesting part of the problem, let's solve it in the end; \item Case 3: s = abac s_l exists on s_l + 1... s_r (can be done using prefix sums), s = baca s_r exists on s_l... s_r - 1 (can be done using prefix sums). To check s = baac we can check, that lt_i ≤ r for some l ≤ i ≤ r, that can be done using minimum on segment in the array lt. \end{itemize}

Now we should the hardest part of this problem~--- we have some segments [l, r]. For all of them, we should check that the border of s[l... r] exists. Here I know two methods, that uses only suffix array. Easiest of them:

We have segment [l, r]. Let's check for all lengths , that s[l... (l + b - 1)] = s[(r - b + 1)... r]. If we don't find border, if it exists, it's length . Let's define i~--- maximal index i such that lcp(l, i) ≥ r - i + 1, and string s[i... r] is a border of s[l... r]. So . But it's easy to see, that the distance between l and i in suffix array , so we need to check only variants of i.

Another method can check that border exists for all segments [l, r] using offline algorithm by O(q·log(n)²) time.

So the total complexity will be or O((n + q)·log(n)²).

#include <bits/stdc++.h>

using namespace std;

const int BIG = 1e9 + 239;
const int M = 2 * 1e5 + 239;
const int L = 19;
const int A = 30;
const int T = (1 << 19);
const int two = 2;

int flm[two][M];

inline void z_function(string &s, int c)
{
	int n = s.length();
	flm[c][0] = 0;
	int l = 0;
	int r = 0;
	for (int i = 1; i < n; i++)
	{
		flm[c][i] = min(flm[c][i - l], r - i);
		if (flm[c][i] < 0) flm[c][i] = 0;
		while (i + flm[c][i] < n && s[flm[c][i]] == s[i + flm[c][i]]) flm[c][i]++;
		if (i + flm[c][i] > r)
		{
			l = i;
			r = i + flm[c][i];
		}
	}        
}

int a[M], lcp[M], pos[M];

inline void suffix_array(string s)
{
	s += (char)(31);
	int n = s.length();
	vector<pair<char, int> > v;
	for (int i = 0; i < n; i++)
		v.push_back(make_pair(s[i], i));
	sort(v.begin(), v.end());
	vector<pair<int, int> > num;
	int last = 0;
	for (int i = 0; i < n - 1; i++)
	{
		num.push_back(make_pair(last, v[i].second));
		if (v[i].first != v[i + 1].first) last++;
	}
	num.push_back(make_pair(last, v.back().second));
	vector<int> u(n);
	for (int i = 0; i < n; i++) u[num[i].second] = num[i].first;
	int d = 1;
	vector<pair<pair<int, int>, int> > t;
	vector<vector<pair<int, int> > > h;
	while (d < n)
	{
		t.clear();
		h.clear();
		h.resize(n);
		for (int i = 0; i < n; i++)
		{
			int l = num[i].second - d;
			if (l < 0) l += n;
			h[u[l]].push_back(make_pair(num[i].first, l));
		}
		for (int i = 0; i < n; i++)
			for (pair<int, int> r : h[i])
				t.push_back(make_pair(make_pair(i, r.first), r.second));
		last = 0;
		num.clear();
		for (int i = 0; i < n - 1; i++)
		{
			num.push_back(make_pair(last, t[i].second));
			if (t[i].first != t[i + 1].first) last++;
		}
		num.push_back(make_pair(last, t.back().second));
		for (int i = 0; i < n; i++) u[num[i].second] = num[i].first;
		d <<= 1;
	}                           
	for (int i = 1; i < n; i++) a[i - 1] = num[i].second;
}

string s;                 

inline void kasai()
{
	int n = s.size();
	suffix_array(s); 
	for (int i = 0; i < n; i++)
		pos[a[i]] = i;
	int k = 0;
	for (int i = 0; i < n; i++)
	{
		if (pos[i] == n - 1) continue;
		while (s[i + k] == s[a[pos[i] + 1] + k] && a[pos[i] + 1] + k < n && i + k < n) k++;
		lcp[pos[i]] = k;
		k = max(0, k - 1);
	}
}

int n, lw[M], rw[M]; 
string prr, rvl;                     
vector<int> open_l[M], close_l[M];   
vector<int> open_r[M], close_r[M];
multiset<int> nw;

inline void func(int l, int r)
{
	if (r - l == 1) return;
	int mid = (l + r) >> 1;
	func(l, mid);
	func(mid, r);
	rvl = "";
	for (int i = mid - 1; i >= l; i--) rvl += s[i];
	z_function(rvl, 0);
	prr = "";
	for (int i = mid; i < r; i++) prr += s[i];
	prr += '#';
	for (int i = l; i < mid; i++) prr += s[i];
	z_function(prr, 1);	
	for (int c = l; c < mid; c++)
	{
		int k1 = 0;
		if (c > l) k1 = flm[0][mid - c];
		int k2 = flm[1][r - mid + 1 + c - l];
		int len = mid - c;
		int lg = max(len - k2, 0);
		int rg = min(len - 1, k1);
		if (rg >= lg)
		{
			open_l[c - rg].push_back((2 * len));
			close_l[c - lg].push_back((2 * len));
			open_r[c - rg + 2 * len - 1].push_back((2 * len));
			close_r[c - lg + 2 * len - 1].push_back((2 * len));
		}	
	}
	rvl = "";
	for (int i = mid; i < r; i++) rvl += s[i];
	z_function(rvl, 0);
	prr = "";
	for (int i = mid - 1; i >= l; i--) prr += s[i];
	prr += '#';
	for (int i = r - 1; i >= mid; i--) prr += s[i];
	z_function(prr, 1); 
	for (int c = mid; c < r; c++)
	{
		int k1 = 0;
		if (c != r - 1) k1 = flm[0][c + 1 - mid];
		int k2 = flm[1][r - c + mid - l];
		int len = c - mid + 1;
		int lg = max(len - k2, 0);
		int rg = min(len - 1, k1);
		if (rg >= lg)
		{                                            
			open_l[c + lg - 2 * len + 1].push_back((2 * len));
			close_l[c + rg - 2 * len + 1].push_back((2 * len));
			open_r[c + lg].push_back((2 * len));
			close_r[c + rg].push_back((2 * len));
		}	
	}
	for (int i = l; i < r; i++)
	{
		for (int len : open_l[i]) nw.insert(len);
		if (!nw.empty()) lw[i] = min(lw[i], *nw.begin());
		for (int len : close_l[i]) nw.erase(nw.lower_bound(len));
		open_l[i].clear();
		close_l[i].clear();
	}          
	for (int i = l; i < r; i++)
	{
		for (int len : open_r[i]) nw.insert(len);
		if (!nw.empty()) rw[i] = min(rw[i], *nw.begin());
		for (int len : close_r[i]) nw.erase(nw.lower_bound(len));
		open_r[i].clear();
		close_r[i].clear();
	}
}

int mn[L][M], st2[M], lc[L][M], kol[A][M];

inline int getmin(int l, int r)
{
	int u = st2[r - l + 1];
	return min(mn[u][l], mn[u][r - (1 << u) + 1]);
}

inline int gett(int l, int r)
{
	if (l == r) return (n - l);
	l = pos[l];
	r = pos[r];
	if (l > r) swap(l, r);
	r--;
	int u = st2[r - l + 1];
	return min(lc[u][l], lc[u][r - (1 << u) + 1]);
}

int q, la[M], ra[M], has[M], mnp[M];

inline void init()
{
	for (int i = 0; i < n; i++) mn[0][i] = lw[i];
	for (int i = 0; i < n - 1; i++) lc[0][i] = lcp[i];
	for (int i = 1; i < L; i++)
		for (int j = 0; j < n; j++)
		{
			int r = (j + (1 << (i - 1)));
			if (r >= n)
			{
				mn[i][j] = mn[i - 1][j];
				continue;
			}
			mn[i][j] = min(mn[i - 1][j], mn[i - 1][r]);
		}
	for (int i = 1; i < L; i++)
		for (int j = 0; j < n - 1; j++)
		{
			int r = (j + (1 << (i - 1)));
			if (r >= n - 1)
			{
				lc[i][j] = lc[i - 1][j];
				continue;
			}                                          
			lc[i][j] = min(lc[i - 1][j], lc[i - 1][r]);
		}	
	st2[1] = 0;
	for (int i = 2; i <= n; i++)
	{
		st2[i] = st2[i - 1];
		if ((1 << (st2[i] + 1)) <= i) st2[i]++;
	}	
}

int in[L][M], gl[T];

inline void build(int i, int l, int r)
{
	if (i == 0) gl[i] = 0;
	else gl[i] = gl[(i - 1) / 2] + 1;
	if (r - l == 1)
	{
		in[gl[i]][l] = a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(2 * i + 1, l, mid);
	build(2 * i + 2, mid, r);
	merge(in[gl[i] + 1] + l, in[gl[i] + 1] + mid, in[gl[i] + 1] + mid, in[gl[i] + 1] + r, in[gl[i]] + l);
}

inline bool is_on(int i, int l, int r, int ql, int qr, int xl, int xr)
{
	if (r <= ql || qr <= l) return false;
	if (ql <= l && r <= qr)
	{
		int it = upper_bound(in[gl[i]] + l, in[gl[i]] + r, xl) - in[gl[i]] - l;
		if (it == r - l) return false;
		return (in[gl[i]][it + l] <= xr);	
	}
	int mid = (l + r) >> 1;
	if (is_on(2 * i + 1, l, mid, ql, qr, xl, xr)) return true;
	return is_on(2 * i + 2, mid, r, ql, qr, xl, xr);
}

inline bool check_all(int l, int r)
{
	int len = (r - l + 1);
	int pl = pos[l];
	int lf = pl;
	int rf = n;
	while (rf - lf > 1)
	{
		int h = (lf + rf) / 2;
		if (gett(a[h], l) >= len) lf = h;
		else rf = h;
	}
	int rg = lf + 1;
	lf = -1;
	rf = pl;
	while (rf - lf > 1)
	{
		int h = (lf + rf) / 2;
		if (gett(a[h], l) >= len) rf = h;
		else lf = h;
	}
	int lg = rf;
	return is_on(0, 0, n, lg, rg, l, r);
}

int par[M];
set<int> cmp[M];
set<pair<int, int> > sc[M];
vector<int> ok[M];
vector<pair<int, int> > mg[M];

inline void border_check()
{
	for (int i = 0; i < q; i++) has[i] = false;
	for (int i = 0; i < q; i++) ok[ra[i] - la[i]].push_back(i);
	for (int i = 0; i < n - 1; i++) mg[lcp[i]].push_back(make_pair(a[i], a[i + 1]));
	for (int i = 0; i < n; i++)
	{
		par[i] = i;
		cmp[i].insert(i);
	}
	for (int c = n; c >= 1; c--)
	{
		for (int i : ok[c - 1])
			sc[par[la[i]]].insert(make_pair(ra[i], i));   
		for (pair<int, int> t : mg[c])
		{
			int l = par[t.first];
			int r = par[t.second];
			if (cmp[l].size() > cmp[r].size()) swap(l, r);
			for (int x : cmp[l])
				while (true)
				{
					auto it = sc[r].lower_bound(make_pair(x, 0));
					if (it == sc[r].end() || it->first > x + c - 1) break;
					has[it->second] = true;
					sc[r].erase(it);
				}
			for (pair<int, int> u : sc[l])
			{
				int id = u.second;
				auto uk = cmp[r].lower_bound(ra[id] - c + 1);
				if (uk != cmp[r].end() && (*uk) <= ra[id]) 
					has[id] = true;
				else 
				    sc[r].insert(u);
			}
			for (int x : cmp[l])
			{
				par[x] = r;
				cmp[r].insert(x);
			}
		}		
	}	
    for (int i = 0; i < q; i++)
	    if (!has[i] && check_all(la[i], ra[i]))
		{
			has[i] = true;
			continue;
		}
}

inline bool checkno(int l, int r)
{
	if (r - l + 1 > 26) return false;
	vector<int> kol(26, 0);
	for (int x = l; x <= r; x++)
	{
		if (kol[s[x] - 'a'] > 0) return false;
		kol[s[x] - 'a']++;
	}	
	return true;
}

inline bool try_kol(int l, int r, int p)
{
	int len = (r - l + 1) / p;
	return (gett(l, l + len) >= (r - l + 1 - len));
}

inline bool ison(int l, int r, char x)
{
	return (kol[x - 'a'][r + 1] > kol[x - 'a'][l]);	
}

inline int query(int l, int r, int id)
{         
	if (checkno(l, r)) return -1;
	int len = (r - l + 1);
	while (len > 1)
	{
		int p = mnp[len];
		if (try_kol(l, r, p)) return 1;
		while ((len % p) == 0) len /= p;
	}
	if (lw[l] <= r) return 2;
	if (rw[r] >= l) return 2;
	if (has[id]) return 2;
	if (ison(l + 1, r, s[l])) return 3;
	if (ison(l, r - 1, s[r])) return 3;
	if (getmin(l, r) <= r) return 3;	
	return 4;
}

int main()
{
	cin.sync_with_stdio(0); 
	cin >> n >> s;
	memset(mnp, -1, sizeof(mnp));
	for (int i = 2; i <= n; i++)
		if (mnp[i] == -1)
			for (int j = i; j <= n; j += i)
				if (mnp[j] == -1)
					mnp[j] = i;
	memset(kol[0], 0, sizeof(kol[0]));
	for (int i = 0; i < n; i++)
	{
		for (int x = 0; x < 26; x++) kol[x][i + 1] = kol[x][i];
		kol[s[i] - 'a'][i + 1]++;
	}
	for (int i = 0; i < n; i++)
	{
		lw[i] = n + 1;
		rw[i] = n + 1;
	}
	func(0, n);  
	for (int i = 0; i < n; i++)
	{
		if (lw[i] == n + 1) lw[i] = n;
		else lw[i] += (i - 1);  
		if (rw[i] == n + 1) rw[i] = -1;
		else rw[i] = (i - rw[i] + 1);
	}
	kasai();                                                                        
	init();  
	build(0, 0, n);                                                                       
	cin >> q;
	for (int i = 0; i < q; i++)
	{
		cin >> la[i] >> ra[i];
		la[i]--, ra[i]--;                       
	}
	border_check();
	for (int i = 0; i < q; i++) cout << query(la[i], ra[i], i) << "\n";
	return 0;
}

#include <bits/stdc++.h>

using namespace std;

#define TIME (clock() * 1.0 / CLOCKS_PER_SEC)

const int BIG = 1e9 + 239;
const int M = 2 * 1e5 + 239;
const int L = 19;
const int A = 30;
const int T = (1 << 19);
const int two = 2;

int flm[two][M];

inline void z_function(string &s, int c)
{
	int n = s.length();
	flm[c][0] = 0;
	int l = 0;
	int r = 0;
	for (int i = 1; i < n; i++)
	{
		flm[c][i] = min(flm[c][i - l], r - i);
		if (flm[c][i] < 0) flm[c][i] = 0;
		while (i + flm[c][i] < n && s[flm[c][i]] == s[i + flm[c][i]]) flm[c][i]++;
		if (i + flm[c][i] > r)
		{
			l = i;
			r = i + flm[c][i];
		}
	}        
}

int a[M], lcp[M], pos[M];

inline void suffix_array(string s)
{
	s += (char)(31);
	int n = s.length();
	vector<pair<char, int> > v;
	for (int i = 0; i < n; i++)
		v.push_back(make_pair(s[i], i));
	sort(v.begin(), v.end());
	vector<pair<int, int> > num;
	int last = 0;
	for (int i = 0; i < n - 1; i++)
	{
		num.push_back(make_pair(last, v[i].second));
		if (v[i].first != v[i + 1].first) last++;
	}
	num.push_back(make_pair(last, v.back().second));
	vector<int> u(n);
	for (int i = 0; i < n; i++) u[num[i].second] = num[i].first;
	int d = 1;
	vector<pair<pair<int, int>, int> > t;
	vector<vector<pair<int, int> > > h;
	while (d < n)
	{
		t.clear();
		h.clear();
		h.resize(n);
		for (int i = 0; i < n; i++)
		{
			int l = num[i].second - d;
			if (l < 0) l += n;
			h[u[l]].push_back(make_pair(num[i].first, l));
		}
		for (int i = 0; i < n; i++)
			for (pair<int, int> r : h[i])
				t.push_back(make_pair(make_pair(i, r.first), r.second));
		last = 0;
		num.clear();
		for (int i = 0; i < n - 1; i++)
		{
			num.push_back(make_pair(last, t[i].second));
			if (t[i].first != t[i + 1].first) last++;
		}
		num.push_back(make_pair(last, t.back().second));
		for (int i = 0; i < n; i++) u[num[i].second] = num[i].first;
		d <<= 1;
	}                           
	for (int i = 1; i < n; i++) a[i - 1] = num[i].second;
}

string s;                 

inline void kasai()
{
	int n = s.size();
	suffix_array(s); 
	for (int i = 0; i < n; i++)
		pos[a[i]] = i;
	int k = 0;
	for (int i = 0; i < n; i++)
	{
		if (pos[i] == n - 1) continue;
		while (s[i + k] == s[a[pos[i] + 1] + k] && a[pos[i] + 1] + k < n && i + k < n) k++;
		lcp[pos[i]] = k;
		k = max(0, k - 1);
	}
}

int n, lw[M], rw[M]; 
string prr, rvl;                     
vector<int> open_l[M], close_l[M];   
vector<int> open_r[M], close_r[M];
multiset<int> nw;

inline void func(int l, int r)
{
	if (r - l == 1) return;
	int mid = (l + r) >> 1;
	func(l, mid);
	func(mid, r);
	rvl = "";
	for (int i = mid - 1; i >= l; i--) rvl += s[i];
	z_function(rvl, 0);
	prr = "";
	for (int i = mid; i < r; i++) prr += s[i];
	prr += '#';
	for (int i = l; i < mid; i++) prr += s[i];
	z_function(prr, 1);	
	for (int c = l; c < mid; c++)
	{
		int k1 = 0;
		if (c > l) k1 = flm[0][mid - c];
		int k2 = flm[1][r - mid + 1 + c - l];
		int len = mid - c;
		int lg = max(len - k2, 0);
		int rg = min(len - 1, k1);
		if (rg >= lg)
		{
			open_l[c - rg].push_back((2 * len));
			close_l[c - lg].push_back((2 * len));
			open_r[c - rg + 2 * len - 1].push_back((2 * len));
			close_r[c - lg + 2 * len - 1].push_back((2 * len));
		}	
	}
	rvl = "";
	for (int i = mid; i < r; i++) rvl += s[i];
	z_function(rvl, 0);
	prr = "";
	for (int i = mid - 1; i >= l; i--) prr += s[i];
	prr += '#';
	for (int i = r - 1; i >= mid; i--) prr += s[i];
	z_function(prr, 1); 
	for (int c = mid; c < r; c++)
	{
		int k1 = 0;
		if (c != r - 1) k1 = flm[0][c + 1 - mid];
		int k2 = flm[1][r - c + mid - l];
		int len = c - mid + 1;
		int lg = max(len - k2, 0);
		int rg = min(len - 1, k1);
		if (rg >= lg)
		{                                            
			open_l[c + lg - 2 * len + 1].push_back((2 * len));
			close_l[c + rg - 2 * len + 1].push_back((2 * len));
			open_r[c + lg].push_back((2 * len));
			close_r[c + rg].push_back((2 * len));
		}	
	}
	for (int i = l; i < r; i++)
	{
		for (int len : open_l[i]) nw.insert(len);
		if (!nw.empty()) lw[i] = min(lw[i], *nw.begin());
		for (int len : close_l[i]) nw.erase(nw.lower_bound(len));
		open_l[i].clear();
		close_l[i].clear();
	}          
	for (int i = l; i < r; i++)
	{
		for (int len : open_r[i]) nw.insert(len);
		if (!nw.empty()) rw[i] = min(rw[i], *nw.begin());
		for (int len : close_r[i]) nw.erase(nw.lower_bound(len));
		open_r[i].clear();
		close_r[i].clear();
	}
}

int mn[L][M], st2[M], lc[L][M], kol[A][M];

inline int getmin(int l, int r)
{
	int u = st2[r - l + 1];
	return min(mn[u][l], mn[u][r - (1 << u) + 1]);
}

inline int gett(int l, int r)
{
	if (l == r) return (n - l);
	l = pos[l];
	r = pos[r];
	if (l > r) swap(l, r);
	r--;
	int u = st2[r - l + 1];
	return min(lc[u][l], lc[u][r - (1 << u) + 1]);
}

int q, mnp[M];

inline void init()
{
	for (int i = 0; i < n; i++) mn[0][i] = lw[i];
	for (int i = 0; i < n - 1; i++) lc[0][i] = lcp[i];
	for (int i = 1; i < L; i++)
		for (int j = 0; j < n; j++)
		{
			int r = (j + (1 << (i - 1)));
			if (r >= n)
			{
				mn[i][j] = mn[i - 1][j];
				continue;
			}
			mn[i][j] = min(mn[i - 1][j], mn[i - 1][r]);
		}
	for (int i = 1; i < L; i++)
		for (int j = 0; j < n - 1; j++)
		{
			int r = (j + (1 << (i - 1)));
			if (r >= n - 1)
			{
				lc[i][j] = lc[i - 1][j];
				continue;
			}                                          
			lc[i][j] = min(lc[i - 1][j], lc[i - 1][r]);
		}	
	st2[1] = 0;
	for (int i = 2; i <= n; i++)
	{
		st2[i] = st2[i - 1];
		if ((1 << (st2[i] + 1)) <= i) st2[i]++;
	}	
}

int u, len;

inline bool check(int l, int r)
{
	for (int t = 1; t <= min(len, ((r - l + 1) / 2)); t++)
		if (gett(l, r - t + 1) >= t)
			return true;
	for (int i = max(0, pos[l] - len); i <= min(n - 1, pos[l] + len); i++)
		if (l < a[i] && a[i] <= r && gett(l, a[i]) >= r - a[i] + 1) 
			return true;
	return false;
}

inline bool checkno(int l, int r)
{
	if (r - l + 1 > 26) return false;
	vector<int> kol(26, 0);
	for (int x = l; x <= r; x++)
	{
		if (kol[s[x] - 'a'] > 0) return false;
		kol[s[x] - 'a']++;
	}	
	return true;
}

inline bool try_kol(int l, int r, int p)
{
	int len = (r - l + 1) / p;
	return (gett(l, l + len) >= (r - l + 1 - len));
}

inline bool ison(int l, int r, char x)
{
	return (kol[x - 'a'][r + 1] > kol[x - 'a'][l]);	
}

inline int query(int l, int r)
{         
	if (checkno(l, r)) return -1;
	int len = (r - l + 1);
	while (len > 1)
	{
		int p = mnp[len];
		if (try_kol(l, r, p)) return 1;
		while ((len % p) == 0) len /= p;
	}
	if (lw[l] <= r) return 2;
	if (rw[r] >= l) return 2;
	if (check(l, r)) return 2;
	if (ison(l + 1, r, s[l])) return 3;
	if (ison(l, r - 1, s[r])) return 3;
	if (getmin(l, r) <= r) return 3;	
	return 4;
}

int main()
{
	cin.sync_with_stdio(0); 
	cin >> n >> s;
	memset(mnp, -1, sizeof(mnp));
	for (int i = 2; i <= n; i++)
		if (mnp[i] == -1)
			for (int j = i; j <= n; j += i)
				if (mnp[j] == -1)
					mnp[j] = i;
	memset(kol[0], 0, sizeof(kol[0]));
	for (int i = 0; i < n; i++)
	{
		for (int x = 0; x < 26; x++) kol[x][i + 1] = kol[x][i];
		kol[s[i] - 'a'][i + 1]++;
	}
	for (int i = 0; i < n; i++)
	{
		lw[i] = n + 1;
		rw[i] = n + 1;
	}
	func(0, n);  
	for (int i = 0; i < n; i++)
	{
		if (lw[i] == n + 1) lw[i] = n;
		else lw[i] += (i - 1);  
		if (rw[i] == n + 1) rw[i] = -1;
		else rw[i] = (i - rw[i] + 1);
	}
	kasai();                                                                        
	init();  
	cin >> q;
	for (int i = 0; i < n; i++)
		if (i * i >= n)
		{
			len = i + 3;
			break;
		}
	for (int i = 0; i < q; i++)
	{
		int l, r;
		cin >> l >> r;
		l--, r--;
		cout << query(l, r) << "\n";                       
	}
	return 0;
}