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<algorithm>
using namespace std;

int n,sz[1000000],par[1000000];
long long d[1000000],chdst[1000000];
pair<long long,int>id[1000000];

bool solve()
{
	for(int i=0;i<n;i++)
		id[i]=make_pair(d[i],i);
	sort(id,id+n);
	for(int i=0;i<n;i++)
	{
		sz[i]=1,par[i]=-2;
		chdst[i]=0;
	}
	for(int i=n-1;i >= 1;--i)
	{
		int at=id[i].second;
		long long want=d[at]-n+2 * sz[at];
		int idx=lower_bound(id,id+n,make_pair(want,-1))-id;
		if(idx >= i || id[idx].first!=want)
			return false;
		par[at]=id[idx].second;
		sz[par[at]] += sz[at];
		chdst[par[at]] += chdst[at]+sz[at];
	}
	int root=id[0].second;
	par[root]=-1;
	return chdst[root]==d[root];
}

int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%lld",&d[i]);
	if(!solve())
	{
		printf("-1");
		return 0;
	}
	for(int i=0;i<n;i++)
		if(par[i]!=-1)
			printf("%d %d\n",i+1,par[i]+1);
	return 0;
}