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>
#define int long long
using namespace std;
int n, sz[100007], fa[100007];
pair<int, int>p[100007];
vector<int>e[100007];
int read()
{
	int num = 0;
	char c = getchar();
	while (c<'0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int w, int f)
{
	int res = 0;
	for (int i = 0, lim = e[w].size(); i < lim; i++)
		res += dfs(e[w][i], w) + sz[e[w][i]];
	return res;
}
signed main()
{
	n = read();
	for (int i = 1; i <= n; i++)
		p[i] = make_pair(read(), i);
	sort(p + 1, p + n + 1);
	for (int i = n; i > 1; i--)
	{
		int x = p[i].second;
		sz[x]++;
		int pos = lower_bound(p + 1, p + n + 1, make_pair(p[i].first + 2 * sz[x] - n, 0LL)) - p;
		if (pos >= i || p[i].first + 2 * sz[x] - n != p[pos].first) { cout << -1 << endl; return 0; }
		fa[x] = p[pos].second;
		sz[fa[x]] += sz[x];
		e[fa[x]].push_back(x);
	}
	sz[p[1].second]++;
	if(dfs(p[1].second,0)!=p[1].first) { cout << -1 << endl; return 0; }
	for (int i = 1; i <= n; i++)
		if (fa[i])printf("%lld %lld\n", fa[i], i);
	return 0;
}