/*
{
######################
#       Author       #
#        Gary        #
#        2020        #
######################
*/
//#pragma GCC target ("avx2")
//#pragma GCC optimization ("O3")
//#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;++a)
#define rl(a,b,c) for(int a=b;a>=c;--a)
#define LL long long
#define IT iterator
#define PB push_back
#define II(a,b) make_pair(a,b)
#define FIR first
#define SEC second
#define FREO freopen("check.out","w",stdout)
#define rep(a,b) for(int a=0;a<b;++a)
#define SRAND mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(a) rng()%a
#define ALL(a) a.begin(),a.end()
#define POB pop_back
#define ff fflush(stdout)
#define fastio ios::sync_with_stdio(false)
#define check_min(a,b) a=min(a,b)
#define check_max(a,b) a=max(a,b)
using namespace std;
const int INF=0x3f3f3f3f;
typedef pair<int,int> mp;
/*}
*/
int n;
bool cmp(mp A,mp B){
	return A.SEC<B.SEC;
}
int main(){
	int a[100000+20];
	cin>>n;
	rb(i,1,n) scanf("%d",&a[i]);
	vector<mp> v;
	int row=n;
	stack<mp> pos;
	vector<mp> pos2;
	rl(i,n,1){
		if(a[i]==1){
			pos.push(II(row,i));
			v.PB(II(row--,i));
		}
	}
	rb(i,1,n){
		if(a[i]==2){
			while(pos.size()&&pos.top().SEC<i) pos.pop();
			if(pos.empty()){
				cout<<-1<<endl;
				return 0;
			}
			v.PB(II(pos.top().FIR,i));
			pos2.PB(II(pos.top().FIR,i));
			pos.pop();
		}
	}
	while(!pos.empty())
		pos2.PB(pos.top()),pos.pop();
	sort(ALL(pos2),cmp);
	reverse(ALL(pos2));
	for(auto it:pos2)
		pos.push(it);
	rl(i,n,1){
		if(a[i]==3){
			while(!pos.empty()&&pos.top().SEC<i) pos.pop();
			if(pos.empty()){
				cout<<-1<<endl;
				return 0;
			}
			v.PB(II(row--,i));
			v.PB(II(row+1,pos.top().SEC));
			int las=i;
			rl(j,i-1,1){
				if(a[j]==3){
					v.PB(II(row--,j));
					v.PB(II(row+1,las));
					las=j;
				}
			}
			break;
		}
	}
	cout<<v.size()<<endl;
	for(auto it:v){
		cout<<it.FIR<<' '<<it.SEC<<endl;
	}
	return 0;
}
/** 程序框架：
  *
  *
  *
  *
  **/

    using namespace std;
     
    typedef long long ll;
    typedef long double ld;
    typedef pair<ll, ll> pll;
     
    #define all(x)                      (x).begin(),(x).end()
    #define X                           first
    #define Y                           second
    #define sep                         ' '
    #define endl                        '\n'
    #define debug(x)                    cerr << #x << ": " <<  x << endl;
     
    ll poww(ll a, ll b, ll md) {
        return (!b ? 1 : (b & 1 ? a * poww(a * a % md, b / 2, md) % md : poww(a * a % md, b / 2, md) % md));
    }
     
    const ll MAXN = 1e6 + 10;
    const ll INF = 8e18;
    const ll MOD = 1e9 + 7; // 998244353; // 1e9 + 9;
     
    set<ll> O, T2, T3;
    set<pll> T2t;
     
    int main() {
    	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    	ll n;
    	cin >> n;
    	vector<ll> v3;
    	for (int i = 1; i <= n; i++) {
    		ll x;
    		cin >> x;
    		if (x == 1) O.insert(i);
    		if (x == 2) T2.insert(i);
    		if (x == 3) T3.insert(i), v3.push_back(i);
    	}
     
    	vector<pll> ans;
    	ll r = n;
    	for (ll e : T2) {
    		if (O.lower_bound(e) == O.end()) return cout << -1 << endl, 0;
    		else {
    			auto it = O.lower_bound(e);
    			ans.push_back({r, e});
    			ans.push_back({r, *it});	
    			O.erase(it);
    		}
     
    		r--;
    	}
     
    	
    	reverse(all(v3));
    	for (ll e : v3) {
    		if (T3.upper_bound(e) == T3.end()) {
    			if (O.lower_bound(e) != O.end()) {
    				auto it = O.lower_bound(e);
    				ans.push_back({r - 1, e});
    				ans.push_back({r - 1, *it});
    				ans.push_back({r, *it});
    				r -= 2;
    				O.erase(it);
    			} else return cout << -1 << endl, 0;
    		} else {
    			auto it = T3.upper_bound(e);
    			ans.push_back({r, e});
    			ans.push_back({r, *it});
    			r--;
    			T3.erase(it);	
    		}
    	}
     
    	for (ll e : O) ans.push_back({r, e}), r--;
    	
    	cout << ans.size() << endl;
    	for (pll e : ans) cout << e.X << sep << e.Y << endl;
    	return 0;
    }

1- I tried to make the columns with number 2 ok by matching them with a column with number one in front of it

2- we can't do anything with the remaining columns with number 2 so if anything was remaining from them, there is no answer

3- let's match the last column with number 3 by a column with number one in front of it witch isn't used while now

4- now we can make the other columns with number 3 ok by matching it with the column 3 in front of it

5- we can easily make the columns with number 1 ok now

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<ll,ll>
#define REP(i,n) for(int i=0;i<n;i++)
const ll maxn=100005;
const ll mod=1e9+7;


int main(){
	fast;	
	ll n,k,i,j;
	cin >> n >> k;
	vector<ll> a(n);
	ll s=0;
	for(i=0;i<n;i++){
		cin >> a[i];
		s+=a[i];
	}
	ll x = s/k;
	ll rem = s%k;
	vector<ll> v(k,0);
	for(i=0;i<rem;++i){
		v[i]=1;
	}
	for(i=0;i<k;++i){
		v[i]+=x;
	}
	ll sum=0;
	for(i=0;i<k;i++){
		sum+=v[i]*v[i];
	}
	cout<<sum<<endl;
	
}