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<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; i++)
#define F first
#define S second
#define int long long
using namespace std;

int n;
pair<int, int> a[200020];
vector<int> ed[200020];
map<int, int> mp;
int sz[200020];
int sum = 0;

void dfs(int now, int dept){
	sum += dept;
	rep(i, ed[now].size()){
		dfs(ed[now][i], dept+1);
	}
}

signed main(){
	cin >> n;
	rep(i, n){
		cin >> a[i].F;
		mp[a[i].F] = i;
		a[i].S = i;
	}
	rep(i, n) sz[i] = 1;
	sort(a, a+n);
	reverse(a, a+n);
	rep(i, n-1){
		int val = a[i].F;
		int to = a[i].F + 2*sz[a[i].S] - n;
		if(!mp.count(to)){
			cout << "-1" << endl;
			return 0;
		}
		int par = mp[to];
		sz[par] += sz[a[i].S];
		ed[par].push_back(a[i].S);
	}
	dfs(a[n-1].S, 0);
	if(sum != a[n-1].F){
		cout << "-1" << endl;
		return 0;
	}
	rep(i, n){
		rep(j, ed[i].size()){
			cout << i+1 << " " << ed[i][j]+1 << endl;
		}
	}
	return 0;
}