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;
const int N=100100;
typedef long long ll;
ll sum[N];int sz[N],n;
map<ll,int>mp;
vector<pair<int,int> >egs;
void gofail(){cout<<-1<<'\n';exit(0);}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i){ll x;cin>>x;mp[x]=i;sz[i]=1;sum[i]=0;}
	for(;!mp.empty();){
		ll u=mp.rbegin()->first;int v=mp.rbegin()->second;mp.erase(u);
		if(sz[v]==n){
			if(sum[v]!=u)gofail();
			sort(egs.begin(),egs.end());
			for(auto&i:egs)cout<<i.first<<' '<<i.second<<'\n';
			return 0;
		}else{
			ll w=u+2*sz[v]-n;
			if(!mp.count(w))gofail();
			int x=mp[w];sz[x]+=sz[v];sum[x]+=sum[v]+sz[v];
			egs.emplace_back(min(v,x),max(v,x));
		}
	}
	return 0;
}