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 <iostream>
#include <cstdio>
#include <algorithm>
#define int long long
#define ll long long
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
const int N = 1e5 + 1000;
int n,num; ll dis,siz[N];
int u[N],to[N];
struct node
{
	ll v; int id;
	int friend operator <(node a,node b){return a.v < b.v;}
}p[N];
signed main()
{
	n = read();
	for(int i = 1;i <= n;i ++) {p[i] = node{read(),i}; siz[i] = 1;}
	sort(p + 1,p + 1 + n);
	for(int i = n;i > 1;i --)
	{
		ll Dis = p[i].v - n + (siz[i] << 1);
		int t = lower_bound(p + 1,p + 1 + n,node{Dis,0}) - p;
		if(p[t].v != Dis) {puts("-1"); return 0;}
		num ++; u[num] = p[i].id; to[num] = p[t].id;
		siz[t] += siz[i]; dis += siz[i]; 
	}
	if(dis != p[1].v) {puts("-1"); return 0;}
	else for(int i = 1;i < n;i ++) cout << u[i] << " " << to[i] << "\n";
}
