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 <queue>
#include <algorithm>
#include <tuple>
using namespace std;

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

pii inp[100005];
int siz[100005];
ll dp[100005];
vector<pii> ans;

int main() {
	int N; scanf("%d", &N);
	for (int i = 1; i <= N; i++) {
		ll x; scanf("%lld", &x);
		inp[i] = {x, i};
		siz[i] = 1;
	}
	sort(inp + 1, inp + N + 1);
	for (int i = N; i >= 2; i--) {
		ll cur; int id;
		tie(cur, id) = inp[i];
		int del = N - 2 * siz[id];
		auto it = *lower_bound(inp + 1, inp + i, pii(cur - del, 0));
		if (it.first != cur - del || del <= 0) return !printf("-1");
		siz[it.second] += siz[id];
		dp[it.second] += dp[id] + siz[id];
		if (id > it.second) swap(id, it.second);
		ans.push_back({id, it.second});
	}
	if (dp[inp[1].second] != inp[1].first) return !printf("-1");
	sort(ans.begin(), ans.end());

	for (pii &x : ans) {
		printf("%lld %d\n", x.first, x.second);
	}
}