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;
vector<int> g[100005];
long long d[100005];
int rd[100005];
int sz[100005];
map<long long,int> mp;
queue<int> q;

int main() {
	ios_base::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		cin >> d[i];
		mp[d[i]] = i;
		sz[i] = 1;
	}
	for (int it = 1; it < n; ++it) {
		int u = mp.rbegin()->second;
		long long need = d[u] - (n - sz[u]) + sz[u];
		if (mp.find(need) == mp.end() || mp[need] == u) {
			return cout << -1 << '\n',0;
		} 
		int v = mp[need];
		g[u].push_back(v), g[v].push_back(u);		
		sz[v] += sz[u];
		mp.erase(mp.find(d[u]));
	}
	for (int i = 1; i <= n; ++i) rd[i] = -1;
	rd[1] = 0; q.push(1);
	while (!q.empty()) {
		int u = q.front(); q.pop();
		for (int v : g[u]) if (rd[v] == -1) {
			rd[v] = rd[u] + 1; q.push(v);
		}
	}
	if (accumulate(rd + 1, rd + 1 + n, 0LL) != d[1]) {
		return cout << -1 << '\n',0;
	}
	for (int u = 1; u <= n; ++u) {
		for (int v : g[u]) if (u < v) {
			cout << u << ' ' << v << '\n';
		}
	} 
}