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>
#define ll long long
#define N 100005
using namespace std;
ll n,ds,d[N],sz[N];map<ll,int>S;pair<int,int>v[N];
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&d[i]),S[d[i]]=i,sz[i]=1;
	sort(d+1,d+n+1);
	for(int i=n;i>1;i--)
    {
		int p=S[d[i]];ll d2=d[i]-(n-sz[p])+sz[p];
		if(S.find(d2)==S.end()){puts("-1");return 0;}
		int q=S[d2];sz[q]+=sz[p];ds+=sz[p];
		v[n-i+1]=make_pair(p,q);
	}
	if(d[1]!=ds){puts("-1");return 0;}
	for(int i=1;i<n;i++) printf("%d %d\n",v[i].first,v[i].second);
}