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 MAX_N 100000
using namespace std;
typedef long long lnt;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n, sz[MAX_N+5]; lnt s[MAX_N+5], tot;
map <lnt, int> id; vector <int> G[MAX_N+5];
void DFS(int u, int d) {tot += d; for (int v : G[u]) DFS(v, d+1);}
int main() {
	read(n);
	for (int i = 1; i <= n; i++)
		read(s[i]), id[s[i]] = i, sz[i] = 1;
	sort(s+1, s+n+1, greater<lnt>());
	for (int i = 1; i < n; i++) {
		int u = id[s[i]];
		lnt sf = s[i]-n+2*sz[u];
		if (sf >= s[i] || !id.count(sf))
			return puts("-1"), 0;
		int f = id[sf];
		sz[f] += sz[u];
		G[f].push_back(u);
	}
	DFS(id[s[n]], 0);
	if (tot != s[n])
		return puts("-1"), 0;
	for (int u = 1; u <= n; u++)
		for (int v : G[u]) printf("%d %d\n", u, v);
	return 0;
}