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>
using namespace std;
const int N=100005;
int main()
{
	ios::sync_with_stdio(false);
	int n;
	cin>>n;
	vector<pair<long long,int>>e;
	map<long long,int>ret;
	for(int i=0;i<n;i++)
	{
		long long x;
		cin>>x;
		ret[x]=i;
		e.push_back({x,i});
	}
	sort(e.rbegin(),e.rend());
	long long x=e.back().first;
	vector<int>sz(n,1);
	vector<long long>dp(n,0);
	vector<pair<int,int>>ans;
	for(auto c:e)
	{
		if(c.first!=x)
		{
			if(sz[c.second]*2>=n)
			{
				cout<<-1<<endl;
				return 0;
			}
			long long need=c.first-(n-2*sz[c.second]);
			if(!ret.count(need))
			{
				cout<<-1<<endl;
				return 0;
			}
			int par=ret[need];
			dp[par]+=dp[c.second]+sz[c.second];
			sz[par]+=sz[c.second];
			ans.push_back({c.second+1,par+1}); 
		}
	}
	if(dp[e.back().second]!=x)
	{
		cout<<-1<<endl;
		return 0;
	}
	for(auto c:ans)
		cout<<c.first<<' '<<c.second<<endl;
	return 0;
}
