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;

int n, cnt[100005];
long long d[100005];
pair<long long, int> p[100005];
vector<int> to[100005];
int ta;
long long tb;

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lld", &p[i].first);
		p[i].second = i + 1;
		cnt[i] = 1;
	}
	sort(p, p + n);
	for (int i = n - 1; i; i--) {
		tb = p[i].first - n + 2 * cnt[i];
		ta = lower_bound(p, p + i, make_pair(tb, 0)) - p;
		if (p[ta].first != tb) {
			printf("-1\n");
			return 0;
		}
		d[ta] += d[i] + cnt[i];
		to[ta].push_back(i);
		cnt[ta] += cnt[i];
	}
	if (d[0] != p[0].first) {
		printf("-1\n");
		return 0;
	}
	for (int i = 0; i < n; i++)
		for (int j : to[i])
			printf("%d %d\n", p[i].second, p[j].second);
	return 0;
}