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>
using namespace std;
typedef bool boolean;

const int N = 1e5 + 5;

#define ll long long
#define par pair<ll, int>

int n;
par a[N];
int Fa[N];
int sz[N], uf[N], d[N];
vector<pair<int, int>> E;

int find(int x) {
	return uf[x] == x ? x : (uf[x] = find(uf[x]));
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i].first);
		a[i].second = i;
		sz[i] = 1;
		uf[i] = i;
	}
	sort(a + 1, a + n + 1);
	for (int i = n; i > 1; i--) {
		ll d = a[i].first;
		int p = a[i].second;
		int fa = lower_bound(a + 1, a + n + 1, par(d - n + (sz[p] << 1), 0)) - a;
		if (a[fa].first != d - n + (sz[p] << 1)) {
			puts("-1");
			return 0;
		}
		if (find(p) == find(fa = a[fa].second)) {
			puts("-1");
			return 0;
		}
		sz[fa] += sz[p];
		E.emplace_back(p, fa);
		Fa[p] = fa;
	}
	ll f1 = 0;
	for (int i = 2; i <= n; i++)
		f1 += (d[a[i].second] = d[Fa[a[i].second]] + 1);
	if (f1 ^ a[1].first) {
		puts("-1");
		return 0;
	}
	for (auto x : E)
		printf("%d %d\n", x.first, x.second);
	return 0;
}
