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 eb emplace_back
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
void fuk() { puts("-1"); exit(0); }

const int MAXN = 100055;

map<ll, int> MP;

vector<pii> EV;

ll A[MAXN];
int B[MAXN], C[MAXN], dep[MAXN];

int N;

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1; i <= N; i++) {
		cin >> A[i];
		MP[A[i]] = i;
	}

	iota(B, B+N+1, 0); sort(B+1, B+N+1, [&](int a, int b) {
		return A[a] < A[b];
	});

	fill(C+1, C+N+1, 1);
	for(int oi = N, i; oi > 1; oi--) {
		i = B[oi];
		ll t = A[i] - N + C[i]*2;
		auto it = MP.find(t);
		if(MP.end() == it) fuk();
		int p = it -> second;
		if(p == i) fuk();
		C[p] += C[i];
		EV.eb(p, i);
	}

	{
		reverse(EV.begin(), EV.end());
		for(auto &e : EV)
			dep[e.second] = dep[e.first] + 1;

		ll sum = 0;
		for(int i = 1; i <= N; i++) sum += dep[i];
		if(sum != A[B[1]]) fuk();
	}

	for(auto &e : EV) printf("%d %d\n", e.first, e.second);
	return 0;
}