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 fi first
#define se second
#define mp make_pair
using namespace std;

typedef pair <long long, int> pli;
typedef pair <int, int> pii;

long long d[100010];
map <long long, int> all;
long long dp[100010], size[100010];

int main(){
	int n; scanf("%d", &n);
	for (int i = 0; i < n; i++){
		scanf("%lld", &d[i]);
		all[d[i]] = i;
	}
	vector <pii> ans;
	int pos = all.begin()->se;
	for (auto it = all.rbegin(); it != all.rend(); it++){
		long long x = it->fi; int id = it->se;
		if (id == pos) continue;
		size[id]++;
		long long need = d[id] + 2 * size[id] - n;
		if (!all.count(need) || need >= x){
			printf("-1\n");
			return 0;
		}
		int fa = all[need];
		ans.push_back(mp(fa, id));
		size[fa] += size[id], dp[fa] += dp[id] + size[id];
	}
	if (dp[pos] != d[pos]){
		printf("-1\n");
		return 0;
	}
	for (int i = 0; i < ans.size(); i++){
		printf("%d %d\n", ans[i].fi + 1, ans[i].se + 1);
	}
	return 0;
}
