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;
typedef long long ll;
const int MAXN = 1e5 + 5;
ll n;
ll d[MAXN], sum;
vector< ll > adj[MAXN];
map< ll , ll > mp;
int h[MAXN], sz[MAXN];
void dfs(int u){
	sum += h[u];
	for(int i = 0; i < adj[u].size(); i++){
		ll v = adj[u][i];
		h[v] = h[u] + 1;
		dfs(v);
	}
	
	return;
}
int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >> n;
	for(int i = 0; i < n; i ++) cin >> d[i], mp[d[i]] = (ll)i;
	for( int i = 0; i < MAXN; i ++) sz[i] = 1LL;
	sort(d, d + n, greater<ll>());
	for(int i = 0; i < n - 1; i ++){
		ll ind = mp[d[i]];
		ll s2 = d[i] + 2LL*sz[ind] - n;
		if( mp.count(s2)){
			if( s2 == d[i]) return cout << -1 << endl, 0;
			ll indpar = mp[s2];
			sz[indpar] += sz[ind], adj[indpar].push_back(ind);
		}
		else return cout << -1 << endl, 0;
	}
	h[mp[d[n - 1]]] = 0;
	dfs(mp[d[n - 1]]);
	if( sum != d[n - 1]) return cout << -1 << endl, 0;
	for(int i = 0; i < n; i ++){
		for( int j = 0; j < adj[i].size(); j ++){
			cout << i + 1LL << " " << adj[i][j] + 1LL << endl;
		}
	}
	
	return 0;
}