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 int long long
using namespace std;

const int N = 100001;
struct node {
	int d, id;
	node() {}
	node(int d, int x):d(d), id(x) {}
	friend bool operator < (node a, node b) { return a.d < b.d; }
}d[N];
int sz[N], u[N], v[N];

inline int read() {
	int x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - 48;
		ch = getchar();
	}
	return x * flag;
}

signed main() {
	int n = read();
	for (int i = 1; i <= n; ++i) {
		d[i] = node{read(), i};
		sz[i] = 1;
	}
	sort(d + 1, d + 1 + n);
	int sum = 0, cnt = 0;
	for (int i = n; i > 1; --i) {
		int dis = d[i].d - n + (sz[i] << 1);
		int pos = lower_bound(d + 1, d + 1 + n, node{dis, 0}) - d;
		if (d[pos].d != dis) return puts("-1"), 0;
		cnt++;
		u[cnt] = d[i].id;
		v[cnt] = d[pos].id;
		sz[pos] += sz[i];
		sum += sz[i];
	}
	if (sum != d[1].d) puts("-1");
	else {
		for (int i = 1; i <= cnt; ++i) {
			printf("%lld %lld\n", u[i], v[i]);
		}
	}
	return 0;
}