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
/*
到所有点距离和最大的一定是叶节点，
而如果i,j相连，一定满足a_j=a_i+2*sz_i-n。
由于做到i的时候i的儿子都做过了，那j一定是i的父亲
然后我们发现还有一个a_i没利用，而这也是所有a_i的最小值 
我们只要判断一下它是不是重心就行了
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=100002;
int a[N],fa[N],id[N],i,sz[N],n,x;
map<int,int>mp;
bool cmp(int x,int y){return a[x]>a[y];}
signed main(){
	scanf("%lld",&n);
	for (i=1;i<=n;i++) scanf("%lld",&a[i]),id[i]=i,mp[a[i]]=i,sz[i]=1;
	sort(id+1,id+n+1,cmp);
	for (i=1;i<n;i++){
		x=id[i];
		fa[x]=mp[a[x]+2*sz[x]-n];
		if (!fa[x]) return puts("-1"),0;
		sz[fa[x]]+=sz[x];
	}
	for (i=1;i<n;i++) a[id[n]]-=sz[id[i]];
	if (a[id[n]]) return puts("-1"),0;
	for (i=1;i<n;i++) printf("%lld %lld\n",id[i],fa[id[i]]);
}