Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

struct Node {
	int parent{-1}, children{1}, id{};
	long long d, distance_sum{};
};

int main()
{
	int n;
	scanf("%d", &n);
	std::vector<Node> nodes(n);
	for (int i = 0; i < n; i++)
	{
		scanf("%lld", &nodes[i].d);
		nodes[i].id = i;
	}
	std::sort(nodes.begin(), nodes.end(),
		[](const Node& a, const Node& b){ return a.d < b.d; }
	);
	for (auto node{nodes.rbegin()}; node + 1 < nodes.rend(); node++)
	{
		auto parent = std::lower_bound(nodes.begin(), nodes.end(), node->d - n + 2 * node->children,
			[](Node &e, long long value){ return e.d < value; }
		);
		if (parent == nodes.end() || parent->d != node->d - n + 2 * node->children || parent->id == node->id)
		{
			printf("-1\n");
			return 0;
		}
		node->parent = parent->id;
		parent->children += node->children;
		parent->distance_sum += node->distance_sum + node->children;
	}
	if (nodes[0].distance_sum != nodes[0].d)
	{
		printf("-1\n");
		return 0;
	}
	for (auto node{nodes.begin() + 1}; node < nodes.end(); node++)
	{
		printf("%d %d\n", node->id + 1, node->parent + 1);
	}
	return 0;
}