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>
using namespace std;
using lint = long long;
using pi = pair<lint, int>;
const int MAXN = 100005;

void my_ass(bool p){
	if(!p){
		puts("-1");
		exit(0);
	}
}

int n, leaf_size[MAXN];
pi a[MAXN];
vector<int> gph[MAXN];

lint get_dist(int x, int d){
	lint ret = d;
	for(auto &i : gph[x]) ret += get_dist(i, d + 1);
	return ret;
}

int main(){
	scanf("%d",&n);
	for(int i=1; i<=n; i++){
		scanf("%lld",&a[i].first);
		a[i].second = i;
	}
	sort(a + 1, a + n + 1);
	fill(leaf_size + 1, leaf_size + n + 1, 1);
	for(int i=n; i>1; i--){
		lint nxt = a[i].first - n + 2 * leaf_size[i];
		int pos = lower_bound(a + 1, a + n + 1, pi(nxt, -1)) - a;
		if(a[pos].first == nxt && pos < i){
			gph[pos].push_back(i);
			leaf_size[pos] += leaf_size[i];
		}
		else my_ass(0);
	}
	my_ass(get_dist(1, 0) == a[1].first);
	for(int i=1; i<=n; i++){
		for(auto &j : gph[i]){
			if(i < j) printf("%d %d\n", a[i].second, a[j].second);
		}
	}
}
