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;
int sz[100005],par[100005];
pair<long long,int> a[100005];
map<long long,int> inv;
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i].first);
		a[i].second=i;
		sz[i]=1;
		inv[a[i].first]=i;
	}
	sort(a+1,a+n+1);
	for (int i=n;i>1;i--)
	{
		if (2*sz[a[i].second]>=n)
		{
			printf("-1");
			return 0;
		}
		int p=inv[a[i].first-n+2*sz[a[i].second]];
		if (!p)
		{
			printf("-1");
			return 0;
		}
		sz[p]+=sz[a[i].second];
		par[a[i].second]=p;
	}
	for (int i=2;i<=n;i++)
	a[1].first-=sz[a[i].second];
	if (a[1].first)
	printf("-1");
	else
	{
		for (int i=2;i<=n;i++)
		printf("%d %d\n",par[a[i].second],a[i].second);
	}
}