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 <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int main() {
	int N;
	cin >> N;
	vector<long long> D(N);
	vector<int> p(N);
	for (int i = 0; i < N; ++i) {
		cin >> D[i];
		p[i] = i;
	}
	sort(p.begin(), p.end(), [&](int i, int j) { return D[i] > D[j]; });
	vector<long long> ds = D;
	for (int i = 0; i < N; ++i) ds[i] *= -1;
	sort(ds.begin(), ds.end());
	vector<int> cnt(N, 1);
	bool ok = true;
	vector<pair<int, int> > edges;
	for (int i = 0; i < N - 1; ++i) {
		long long tar = -ds[i] - (N - 2 * cnt[p[i]]);
		int u = lower_bound(ds.begin(), ds.end(), -tar) - ds.begin();
		if (u == N || -ds[u] != tar) {
			ok = false;
			break;
		}
		cnt[p[u]] += cnt[p[i]];
		edges.push_back(make_pair(p[u], p[i]));
	}
	long long sum = 0;
	for (int i = 0; i < N; ++i) {
		if (i != p[N - 1]) {
			sum += cnt[i];
		}
	}
	if (sum != -ds[N - 1]) ok = false;
	if (!ok) {
		cout << -1 << endl;
	}
	else {
		for (pair<int, int> i : edges) {
			cout << i.first + 1 << ' ' << i.second + 1 << endl;
		}
	}
	return 0;
}