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 fi first
#define se second
#define mp make_pair
using namespace std;

int size[100010], fa[100010];
long long dp[100010];
map <long long, int> to_i;
long long D[100010];
int n;

int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &D[i]);
		to_i[D[i]] = i;
		size[i] = 1, dp[i] = 0;
	}
	map <long long, int> :: iterator it = to_i.end();
	while ((--it) != to_i.begin()) {
		int x = it->se;
		if (2 * size[x] >= n || !to_i.count(D[x] + 2 * size[x] - n)) {
			printf("-1\n");
			return 0;
		}
		fa[x] = to_i[D[x] + 2 * size[x] - n];
		size[fa[x]] += size[x], dp[fa[x]] += dp[x] + size[x];
	}
	if (to_i.begin()->fi != dp[to_i.begin()->se]) {
		printf("-1\n");
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		if (fa[i]) printf("%d %d\n", i, fa[i]);
	}
	return 0;
}
