#include <bits/stdc++.h>
using namespace std;
int main(){
	int a, b;
	scanf("%d%d", &a, &b);
	printf("%d\n", a >= b);
}

#include <bits/stdc++.h>
using namespace std;

bool isp(int x){
	if (x < 2)return false;
	for (int i = 2; i * i <= x; ++i)if (x % i == 0)return false;
	return true;
}

int main(){
	int x;
	scanf("%d", &x);
	if (isp(x - 2))printf("%d %d\n", 2, x - 2);
	else printf("-1\n");
}

#include <bits/stdc++.h>


using namespace std;

int main(){

  int s , x ; 
  scanf("%d %d" , &s , &x) ; 
  int ans = 0 ; 
  while (s > 0){
    	s/=x ; 
    	ans++;
  }
  
  if (ans != 0)
    printf("%d\n" , ans) ; 
  else
    printf("%d\n" , 1) ; 

    
}

#include <bits/stdc++.h>
using namespace std;

void solve(){
	string an;
	for (int i = 0; i < 24; ++i)an += "DL";
	an += "RRUU";
	printf("%d\n%s\n", (int)an.size(), an.c_str());
}

int main(){
	int t;
	scanf("%d", &t);
	while (t--)solve();
}

#include <bits/stdc++.h>
using namespace std;
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

void solve(){
	string an;
	f(it, 0, 24)an += "DL";
	f(it, 0, 4){
		f(it, 0, 5)an += 'U';
		f(it, 0, 2)an += 'L';
		f(it, 0, 5)an += 'D';
		an += 'L';
		f(it, 0, 5)an += 'D';
	}
	an += "RRRRRRRRDDUUURDDDDUUUUULLDDDDRRUUUU";
	printf("%d\n%s\n", (int)an.size(), an.c_str());
}

int main(){
	int t;
	scanf("%d", &t);
	while (t--)solve();
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000;
#define pb push_back
int n,x[N],y[N],l[N],r[N],indeg[N];
bool g[N][N];

void get(int &n, int l, int r){
	assert(scanf("%d",&n)+1);
	assert(n>=l && n<=r);
}

set<pair<int, int>> st;
int main(){
	get(n,1,3000);
	for(int a,rn,i=0; i<n; ++i){
		get(x[i], -oo, oo);
		get(y[i], -oo, oo);
		get(a, 0, 359);
		get(rn, 0, 89);
		l[i]=(a-rn+360)%360;
		r[i]=(a+rn)%360;
		assert(st.insert({x[i],y[i]}).second);
	}
	for(int i=0; i<n; ++i){
		for(int j=0; j<n; ++j){
			if(i==j)continue;
			int dx=x[j]-x[i], dy=y[j]-y[i];
			double ang=atan2(dy,dx);
			if(ang<0)
				ang+=2*PI;
			ang*=180/PI;
			bool in=0;
			if(l[i]>r[i]){
				if(ang+eps>=l[i] || ang-eps<=r[i])
					in=1;
			}else if(ang+eps>=l[i] && ang-eps<=r[i])
				in=1;
			g[i][j]=in;
			indeg[j]+=in;
		}
	}
	queue<int> q;
	for(int i=0; i<n; ++i)
		if(!indeg[i])
			q.push(i);

	vector<int> ans;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		ans.push_back(u+1);
		for(int i=0; i<n; ++i)
			if(g[u][i] && !--indeg[i])
				q.push(i);
	}
	if(ans.size()<n)
		return cout<<-1<<endl,0;
	for(int i=0; i<n; ++i)
		printf("%d%c", ans[i], i<n-1?' ':'\n');
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 100000;
ll const inf = 2e18;
int n;

struct S{
	int fr;
	ll a, b, mn;
	S(){}
	S(int _a, ll _b):fr(1), a(_a), b(_b), mn(_a + _b) {}
	S operator +(S const &o){
		S an;
		an.fr = fr + o.fr;
		an.a = a + o.a;
		an.mn = min(mn, o.mn + a);
		return an;
	}
}s[N << 2];

void st(int i, int a1, ll b1, int l = 1, int r = n, int id = 1){
	if (l == r)return void(s[id] = S(a1, b1));
	int m = l + r >> 1, a = id << 1, b = a | 1;
	if (i <= m)st(i, a1, b1, l, m, a);
	else st(i, a1, b1, m + 1, r, b);
	s[id] = s[a] + s[b];
}

void kl(ll x, ll sm = 0, int l = 1, int r = n, int id = 1){
	if (l == r){
		if(sm + s[id].mn >= x)return;
		s[id].fr = 0;
		s[id].a = 0;
		s[id].b = inf;
		s[id].mn = inf;
		return;
	}
	int m = l + r >> 1, a = id << 1, b = a | 1;
	if (s[b].mn + sm + s[a].a < x)kl(x, sm + s[a].a, m + 1, r, b);
	if (s[a].mn + sm < x)kl(x, sm, l, m, a);
	s[id] = s[a] + s[b];
}

int bs(ll x, ll sm = 0, int l = 1, int r = n, int id = 1){
	if (l == r)return x >= sm + s[id].a ? s[id].fr : 0;
	int m = l + r >> 1, a = id << 1, b = a | 1;
	if (s[a].a + sm > x)return bs(x, sm, l, m, a);
	else return s[a].fr + bs(x, sm + s[a].a, m + 1, r, b);
}

int dead(ll x){
	int fr = s[1].fr;
	kl(x);
	return fr - s[1].fr;
}

int hungry(ll x) { return s[1].fr - bs(x); }

int main(){
	scanf("%d", &n);
	f(i, 1, n + 1){
		int a;
		ll b;
		scanf("%d%lld", &a, &b);
		st(i, a, b);
	}
	int q;
	scanf("%d", &q);
	while (q--){
		int o;
		scanf("%d", &o);
		if (o == 1){
			ll x;
			scanf("%lld", &x);
			int b = hungry(x), a = dead(x);
			printf("%d %d\n", a, b);
		}else {
			int a, c;
			ll b;
			scanf("%d%lld%d", &a, &b, &c);
			st(c, a, b);
		}
	}
}

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1) ;  
int main() {
	double v , s ; 
	scanf("%lf %lf" ,&v , &s) ; 
	double ans ; 
	ans = s/(2.0 * sin(pi/v)) ; 
	printf("%.9lf" , ans * ans * pi) ; 
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 1000000;
char s[N + 1];
int an[N], n;

inline bool isNumberDigit(char c) { return c >= '0' && c <= '9'; }

int I;

inline int readInteger(){
	int an = 0;
	while (isNumberDigit(s[I])){
		an = an * 10 + s[I] - '0';
		++I;
	}
	return an;
}

void readSoldier(int supervisor){
	int id = readInteger();
	while (s[I] == '('){
		++I;
		readSoldier(id);
		++I;
	}
	an[id] = supervisor;
}

int main(){
	scanf("%d%s", &n, s);
	
	readSoldier(0);
	
	printf("%d", an[1]);
	f(i, 2, n + 1)printf(" %d", an[i]);
	printf("\n");
}

#include <bits/stdc++.h>
using namespace std;
const int N = 2001,M = 1e9 + 7;
int n,m,k,dp[N][N],ans;
void add(int &a,int b){
	a+=b;
	if(a >= M)a-=M;
}
int main(){
	scanf("%d%d%d",&n,&k,&m);
	k = min(m,k);
	dp[0][0] = 1;
	for(int i = 0;i < m;i++)
		for(int j = 0;j <= k;j++){
			if(j + 1 <= k)add(dp[i + 1][j + 1],dp[i][j]);
			if(j - 1 >= 0)add(dp[i + 1][j - 1],dp[i][j]);
		}
	for(int i = 2;i <= m;i+=2)add(ans,dp[i][0]);
	printf("%d\n",2LL*n*ans%M);
}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll ; 
int n  , a[25];
vector<int>v ; 
ll ans = 0 ; 
void calc(int idx) {
	if (idx == n)
		{
			int x = 0 ; 
			for(int i =0 ; i<v.size() ; ++i) {
				x |=v[i] ; 
			}
			ans+= 1ll *x ; 
			return ; 
		}
		
	v.push_back(a[idx]) ; 
	calc(idx+1) ; 
	
	v.pop_back() ; 
	calc(idx+1) ; 
} 

int main() {

	scanf("%d" , &n) ; 
	for (int i =0 ; i<n ; ++i) {
		scanf("%d" , &a[i]) ; 
	}
	calc(0) ; 
	printf("%lld\n" , ans) ; 
	return 0;
}

Instead of going through all the subsets, we will try to find a simpler way by going through the bits of the given numbers. we know that the number of subsets is $$$2^n - 1$$$, for the $$$i^{th}$$$ bit of each number we know it's either $$$1$$$ or $$$0$$$, you can notice that for a resulting subset of ORing some numbers, its $$$i^{th}$$$ bit is $$$0$$$ $$$iff$$$ all the $$$i^{th}$$$ bit of elements in the subset is $$$0$$$. With this observation we can eliminate all the $$$i^{th}$$$ bits of $$$0's$$$ in all subsets and add keep the $$$1's$$$ and then add the summation of the $$$1's$$$ of $$$i^{th}$$$ bit to the final answer. In short, what you have to do is to find the summation for each $$$i^{th}$$$ bit in all subsets, but don't forget to multiply by its digit place, and you get the resulting formula for the $$$i^{th}$$$ bit (assuming $$$x = 0's$$$ $$$count$$$ $$$for$$$ $$$i^{th}$$$ $$$bit$$$) the answer is : $$$((2^n -1) - (2^x-1)) * 2^i.$$$

Complexity: $$$O(n)$$$

#include <bits/stdc++.h>


using namespace std;
typedef long long ll ; 
const int N = 1e5+5 ; 
int n , a[N] , zero[31]; 
int main() {
	scanf("%d" , &n) ; 
	for (int i =0 ; i<n; ++i) {
		scanf("%d" , &a[i]) ; 
	}


	for (int i = 0; i <31 ; i++){   
           for (int j = 0; j < n; j++) {        
               if (!(a[j] & 1 << i)) {
                   zero[i]++;          
               }
            }
	}
	
    ll ans = 0; 
    for (int i = 0; i < 31 ; i++)  
    { 
        ans += (1ll * ( (1<<n) - 1) -  (1ll*(1<<zero[i]) - 1)) * 1ll*(1<<i); 
    } 
    printf("%lld\n" , ans) ; 
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 200000;
char s[N + 1];
vector<pair<int, int> > an;

void no() { printf("-1\n"); exit(0); }

inline void ad(int tp, int i) { an.push_back(make_pair(tp, i)); }

int main(){
	scanf("%s", s);
	int n = strlen(s);
	string v;
	f(i, 0, n)if (s[i] == 'c'){ // first remove all 'c' characters
		if (v.empty())no(); // 'c' is first character in string
		if (v.back() == 'a'){ // "ac"
			ad(3, v.size() + 1);
			ad(2, v.size() + 1);
			ad(4, v.size());
		}else {
			if (v.size() == 1)no(); // "bc"
			if (v[v.size() - 2] == 'a'){ // "abc"
				ad(4, v.size() - 1);
				v.pop_back();
				v.pop_back();
			}else { // "bbc"
				ad(2, v.size() - 1);
				ad(3, v.size());
				ad(4, v.size() + 1);
			}
		}
	}else v += s[i];
	string g;
	f(i, 0, v.size())if (v[i] == 'b'){ // string contains only 'a' and 'b'
		if (g.empty())no();
		ad(2, g.size() + 1);
		ad(4, g.size());
		g.pop_back();
	}else g += v[i];
	f(i, 0, g.size()){ // remove any left 'a' characters
		ad(1, 1);
		ad(2, 2);
		ad(4, 1);
	}
	printf("%d\n", (int)an.size());
	f(i, 0, an.size())printf("%d %d\n", an[i].first, an[i].second);
}