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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
int n,size[100005],sa[100005],pa[100005];
ll d[100005];
bool cmp(int a,int b)
{
	return d[a]>d[b];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&d[i]);
		sa[i]=i;
	}
	sort(sa+1,sa+n+1,cmp);
	for(int i=1;i<n;i++)
	{
		int x=sa[i];
		size[x]++;
		ll a=d[x]-n+2*size[x];
		int l=i+1,r=n;
		while(l<=r)
		{
			int mid=(l+r)>>1;
			if(d[sa[mid]]<a)r=mid-1;
			else l=mid+1;
		}
		if(d[sa[r]]!=a)
		{
			printf("-1\n");
			return 0;
		}
		pa[x]=sa[r];
		size[sa[r]]+=size[x];
	}
	ll sum=0;
	for(int i=1;i<n;i++)sum+=size[sa[i]];
	if(sum!=d[sa[n]])
	{
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=n;i++)
	  if(pa[i])printf("%d %d\n",i,pa[i]);
	return 0;
}