You are given a sequence D_1, D_2, ..., D_N of length N. The values of D_i are all distinct. Does a tree with N vertices that satisfies the following conditions exist?

* The vertices are numbered 1,2,..., N.
* The edges are numbered 1,2,..., N-1, and Edge i connects Vertex u_i and v_i.
* For each vertex i, the sum of the distances from i to the other vertices is D_i, assuming that the length of each edge is 1.



If such a tree exists, construct one such tree.

Constraints

* 2 \leq N \leq 100000
* 1 \leq D_i \leq 10^{12}
* D_i are all distinct.

Input

Input is given from Standard Input in the following format:


N
D_1
D_2
:
D_N


Output

If a tree with n vertices that satisfies the conditions does not exist, print `-1`.

If a tree with n vertices that satisfies the conditions exist, print n-1 lines. The i-th line should contain u_i and v_i with a space in between. If there are multiple trees that satisfy the conditions, any such tree will be accepted.

Examples

Input

7
10
15
13
18
11
14
19


Output

1 2
1 3
1 5
3 4
5 6
6 7


Input

2
1
2


Output

-1


Input

15
57
62
47
45
42
74
90
75
54
50
66
63
77
87
51


Output

1 10
1 11
2 8
2 15
3 5
3 9
4 5
4 10
5 15
6 12
6 14
7 13
9 12
11 13
#include <iostream>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
int N;
ll D[100010];
vector<pair<int,int>> ans;
vector<vector<int>> v(100010);
ll dist[100010] = {0},inf = 1e18;

void dfs(int n){
	for(int i=0;i<v[n].size();i++){
		int s = v[n][i];
		if(dist[s]==inf){
			dist[s] = dist[n]+1;
			dfs(s);
		}
	}
}

int main(){
	cin >> N;
	map<ll,int> m,num;
	priority_queue<ll> Q;
	for(int i=1;i<=N;i++){
		cin >> D[i];
		m[D[i]] = i;
		num[D[i]] = 1;
		Q.push(D[i]);
		dist[i] = inf;
	}
	int count = 0;
	while(count<N-1 && !Q.empty()){
		ll s = Q.top();
		Q.pop();
		ll t = s-(N-2*num[s]);
		if(t<=0) continue;
		if(m.count(t)){
			ans.push_back({m[s],m[t]});
			v[m[s]].push_back(m[t]);
			v[m[t]].push_back(m[s]);
			count++;
			num[t] += num[s];
		}
	}
	dist[1] = 0;
	dfs(1);
	ll sum = 0;
	for(int i=2;i<=N;i++) sum += dist[i];
	if(count!=N-1 || sum!=D[1]) cout << -1 << endl;
	else{
		for(int i=0;i<N-1;i++){
			cout << ans[i].first << " " << ans[i].second << endl;
		}
	}
}