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 int long long
using namespace std;
const int maxn=1e5+9;
struct node{
	int d,id;
	node(int d=0,int id=0):d(d),id(id){}
	friend bool operator<(node x,node y){
		return x.d<y.d;
	}
}d[maxn];
int n,dis,tot,siz[maxn],u[maxn],v[maxn];
signed main(){
	cin>>n;
	for(int i=1,x;i<=n;++i)cin>>x,d[i]=node(x,i),siz[i]=1;
	sort(d+1,d+1+n);
	for(int i=n;i>1;--i){//最大的一定是叶子 可以递推siz
		int nxt=d[i].d-n+2*siz[i];//下一个的d
		int p=lower_bound(d+1,d+1+n,node(nxt,0))-d;
		if(d[p].d!=nxt)return puts("-1"),0;//如果找不到 不符合
		u[++tot]=d[i].id;v[tot]=d[p].id;
		siz[p]+=siz[i];
		dis+=siz[i];
	}
	if(dis!=d[1].d)puts("-1");//因为只判断了d之差的关系 要看找出的siz是否符合条件
	else{
		for(int i=1;i<=tot;++i)cout<<u[i]<<" "<<v[i]<<endl;
	}
	return 0;
}