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 rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int n;
map<LL, PII> S;
vector<PII> ans;
LL offset;

int main()
{
	scanf("%d", &n);
	rep(i, n) {
		LL d;
		scanf("%lld", &d);
		S[d] = MP(1, i + 1);
	}
	while(S.size() >= 2) {
		pair<LL, PII> cur = *S.rbegin();
		S.erase(cur.first);
		LL nd = cur.first - n + 2 * cur.second.first;
		if(S.find(nd) == S.end()) {
			printf("-1\n");
			return 0;
		}
		S[nd].first += cur.second.first;
		offset += cur.second.first;
		ans.push_back(MP(cur.second.second, S[nd].second));
	}
	if(S.begin()->first != offset) printf("-1\n");
	else rep(i, ans.size()) printf("%d %d\n", ans[i].first, ans[i].second);
	return 0;
}