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 <algorithm>
#include <utility>
#include <cstdio>
#include <map>
using ll = long long;
int fa[100005], sz[100005];
ll arr[100005], val[100005];
std::pair<ll, int> app[100005];
std::map<ll, int> idx;
int main()
{
	// freopen("ARC103-F.in", "r", stdin);
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%lld", arr + i);
		idx[arr[i]] = i;
		app[i] = {arr[i], i};
		sz[i] = 1;
	}
	std::sort(app, app + n);
	for (int i = n - 1; i; i--)
	{
		int u = app[i].second;
		ll dt = n - sz[u] * 2;
		if (dt <= 0 || !idx.count(arr[u] - dt))
		{
			puts("-1");
			return 0;
		}
		fa[u] = idx[arr[u] - dt];
		sz[fa[u]] += sz[u];
		val[fa[u]] += val[u] + sz[u];
	}
	if (val[app[0].second] != arr[app[0].second])
	{
		puts("-1");
		return 0;
	}
	for (int i = n - 1; i; i--)
	{
		int u = app[i].second;
		printf("%d %d\n", fa[u] + 1, u + 1);
	}
	return 0;
}
