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 <algorithm>
#include <vector>
#include <map>
using namespace std;

const int N = 100 * 1000 + 5;

map <long long, int> mp;
vector <pair <int, int> > e;
pair <long long, int> d[N];
int c[N];
long long sum[N];


int main() {
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> d[i].first;
		d[i].second = i;
		c[i] = 1;
	}
	sort(d + 1, d + n + 1);
	for (int i = 1; i <= n; i++) {
		mp[d[i].first] = i;
	}
	for (int i = n; i > 1; i--) {
		int u = mp[d[i].first - n + 2 * c[i]];
	//	cout << "73 " << d[i].second << " " << d[u].second << endl;
		if (u == 0 || u == i) {
			cout << -1;
			return 0;
		} 
		c[u] += c[i];
		sum[u] += c[i] + sum[i];
		e.push_back({d[i].second, d[u].second});
	}
	if (sum[1] != d[1].first) {
		cout << -1;
		return 0;
	} 
	for (auto p : e) {
		cout << p.first << " " << p.second << endl;
	}
}
