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<cstdio>
#include<algorithm>
#include<vector>
#define N_ 101000
using namespace std;
int n, C[N_];
vector<int>E[N_];
struct point {
	long long x;
	int num;
	bool operator<(const point &p)const {
		return x != p.x ? x < p.x : num < p.num;
	}
}w[N_];
long long sum;
void DFS(int a, int d) {
	sum += d;
	for (auto &x : E[a])DFS(x, d + 1);
}
int main() {
	int i;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%lld", &w[i].x);
		w[i].num = i, C[i] = 1;
	}
	sort(w + 1, w + n + 1);
	for (i = n; i > 1; i--) {
		int t = w[i].num;
		long long nx = w[i].x - n + C[t] * 2;
		point tp = { nx,-1 };
		int j =  lower_bound(w + 1, w + n + 1, tp) - w; 
		if (nx >= w[i].x || j<1 || j>n || w[j].x != nx) {
			puts("-1");
			return 0;
		}
		C[w[j].num] += C[t];
		E[w[j].num].push_back(t);
	}
	DFS(w[1].num, 0);
	if (sum != w[1].x)puts("-1");
	else {
		for (i = 1; i <= n; i++)for (auto &x : E[i])printf("%d %d\n", i, x);
	}
}