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<stdio.h>
#include<algorithm>
using namespace std;
struct node{
	int u;
	long long d;
	bool operator<(node b) const
	{
		return d<b.d;
	}
};
node nn[131072];
int size[131072],fa[131072],d[131072];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		nn[i].u=i;
		scanf("%lld",&nn[i].d);
		size[i]=1;
	}
	sort(nn+1,nn+n+1);
	for(int i=n;i>1;i--)
	{
		int l=1,r=i-1;
		while(l<=r)
		{
			int mid=(l+r)>>1;
			if(nn[mid].d==nn[i].d+size[i]-(n-size[i]))
			{
				fa[i]=mid;
				break;
			}
			if(nn[mid].d>nn[i].d+size[i]-(n-size[i]))
			{
				r=mid-1;
			}
			else
			{
				l=mid+1;
			}
		}
		if(!fa[i])
		{
			printf("-1\n");
			return 0;
		}
		size[fa[i]]+=size[i];
	}
	long long sd=0;
	for(int i=2;i<=n;i++)
	{
		d[i]=d[fa[i]]+1;
		sd+=d[i];
	}
	if(sd!=nn[1].d)
	{
		printf("-1\n");
		return 0;
	}
	for(int i=2;i<=n;i++)
	{
		printf("%d %d\n",nn[fa[i]].u,nn[i].u);
	}
	return 0;
}