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<cstring>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
map<long long,int> id;
int size[100005];
long long w[100005];
vector<pair<int,int> >e;
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		long long D;
		scanf("%lld",&D);
		id[D]=i;
	}
	for(int i=1;i<=n;i++)size[i]=1,w[i]=0;
	for(int j=1;j<n;j++){
		long long d=id.rbegin()->first;
		int x=id.rbegin()->second;
		id.erase(d);
		d-=n-2*size[x];
		if(!id.count(d)||size[x]*2>n){puts("-1"); return 0;}
		e.push_back(make_pair(id[d],x));
		size[id[d]]+=size[x];
		w[id[d]]+=w[x]+size[x];
	}
	if(size[id.begin()->second]!=n||w[id.begin()->second]!=id.begin()->first){puts("-1"); return 0;}
	for(int i=0;i<e.size();i++)printf("%d %d\n",e[i].first,e[i].second);
	return 0;
}