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;
typedef long long LL;
const int N=100100;
struct Node{
	LL d;
	LL td;
	int i,sz,fa;
	vector<int> ch;
	Node():td(0),sz(1){}
	bool operator<(const Node o)const{
		return d<o.d;
	}
};
Node d[N];
map<LL,int> fd;
void GG(){
	puts("-1");
	exit(0);
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld",&d[i].d);
		d[i].i=i;
	}
	sort(d+1,d+n+1);
	for(int i=1;i<=n;i++){
		fd[d[i].d]=i;
	}
	for(int i=n;i>1;i--){
		fd.erase(d[i].d);
		LL nd=d[i].d-(n-2*d[i].sz);
		if(!fd.count(nd)){
			GG();
		}
		int f=fd[nd];
		d[i].fa=d[f].i;
		d[f].sz+=d[i].sz;
		d[f].td+=d[i].td+d[i].sz;
	}
	if(d[1].td!=d[1].d){
		GG();
	}
	for(int i=2;i<=n;i++){
		printf("%d %d\n",d[i].fa,d[i].i);
	}
	return 0;
}