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 <map>
#include <vector>
#include <algorithm>

typedef long long int64;

const int maxn = 100005;
int n, size[maxn];
int64 d[maxn];
std::map<int64, int> s;

int main()
{
	static int id[maxn];
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%lld", d + i), s[d[i]] = i, id[i] = i;
	std::sort(id + 1, id + 1 + n, [](int x, int y) { return d[x] < d[y]; });
	std::vector<std::pair<int, int>> e;
	for (int i = 1; i <= n; ++i)
		size[i] = 1;
	for (int i = n; i != 1; --i)
	{
		if (2 * size[id[i]] >= n || !s.count(d[id[i]] - (n - 2 * size[id[i]])))
		{
			puts("-1");
			return 0;
		}
		int fa = s[d[id[i]] - (n - 2 * size[id[i]])];
		e.emplace_back(fa, id[i]);
		size[fa] += size[id[i]];
		d[id[1]] -= size[id[i]];
	}
	if (d[id[1]])
		puts("-1");
	else
		for (const auto &p : e)
			printf("%d %d\n", p.first, p.second);
	return 0;
}
