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;
 
const int N = 100010;
typedef long long int64;
 
map<int64 , int> val;
int n , size[N] , fa[N];
 
int64 d[N];
typedef pair<int64 , int> pii;
priority_queue<pii> leaf;
#define mp make_pair 
 
int main() {
    cin >> n;
    for(int i = 1 ; i <= n ; ++i) {
        cin >> d[i];
        size[i] = 1;
        val[d[i]] = i; 
    }
    for(int i = 1 ; i <= n ; ++i) leaf.push(mp(d[i] , i));
	
	int64 root = 0;
    while(leaf.size()) {
        int x = leaf.top().second;
        leaf.pop();
 
		val.erase(d[x]);
        if(val.count(d[x] - n + size[x]*2))
            fa[x] = val[d[x] - n + size[x]*2];
        else break;
        
        size[fa[x]] += size[x];
        root += size[x];
    }
 
    for(int i = 1 , cnt = 0; i <= n ; ++i) {
    	if(fa[i] == 0 && (++cnt > 1 || d[i] != root))  {cout << -1 << endl; return 0;}
	}
    for(int i = 1 ; i <= n ; ++i)
        if(fa[i]) cout << i << ' ' << fa[i] << endl;
	
    return 0;
}