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>

using namespace std;

int N;
map<long long,int> id;
long long seq[100050];
int sz[100050];
vector<int> v[100050];
long long ans = 0;

void Serval(int pos,int dep,int lst)
{
	ans += dep;
	for(int i = 0;i < v[pos].size(); ++ i)
	{
		int nxt = v[pos][i];
		if(nxt != lst)
			Serval(nxt,dep + 1,pos);
	}
}

int main()
{
	scanf("%d",&N);
	for(int i = 1;i <= N; ++ i)
		scanf("%lld",&seq[i]),id[seq[i]] = i,sz[i] = 1;
	sort(seq + 1,seq + 1 + N);
	for(int i = N;i >= 2; -- i)
	{
		int delta = N - 2 * sz[id[seq[i]]];
		if(delta < 0)
		{
			printf("-1\n");
			return 0;
		}
		long long to_find = seq[i] - delta;
		if(!id.count(to_find))
		{
			printf("-1\n");
			return 0;
		}
		sz[id[to_find]] += sz[id[seq[i]]];
		v[id[seq[i]]].push_back(id[to_find]);
		v[id[to_find]].push_back(id[seq[i]]);
	}
	Serval(id[seq[1]],0,0);
	if(ans != seq[1])
	{
		printf("-1\n");
		return 0;
	}
	for(int i = 1;i <= N; ++ i)
	{
		for(int j = 0;j < v[i].size(); ++ j)
		{
			int nxt = v[i][j];
			if(nxt > i) printf("%d %d\n",i,nxt);
		}
	}
}