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>
#define int long long
using namespace std;
const int N=1e5+2;
map<int,int> idx;
vector<pair<int,int> > ans;
int n,ar[N],sz[N],sz1[N],ar1[N];
vector<int> adj[N];
void dfs(int x,int p){
	sz[x]=1;
	for(int i:adj[x]){
		if(i!=p){
			dfs(i,x);
			sz[x]+=sz[i];
			sz1[x]+=sz1[i]+sz[i];
		}
	}
}
void dfs1(int x,int p){
	for(int i:adj[x]){
		if(i!=p){
			sz1[i]=sz1[x]+n-2*sz[i];
			dfs1(i,x);
		}
	}
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int m,i,j,k,l;
	cin>>n;
	for(i=1;i<=n;i++){
		cin>>ar[i];
		ar1[i]=ar[i];
		idx[ar[i]]=i;
		sz[i]=1;
	}
	sort(ar+1,ar+1+n);
	for(i=n;i>1;i--){
		k=idx[ar[i]]; j=ar[i]+2*sz[k]-n; l=idx[j];
		if(!l||j>=ar[i]){
			cout<<"-1";
			return 0;
		}
		adj[l].push_back(k);
		adj[k].push_back(l);
		sz[l]+=sz[k];
		ans.push_back({k,l});
	}
	dfs(1,1);
	dfs1(1,1);
	for(i=1;i<=n;i++){
		if(sz1[i]!=ar1[i]){
			cout<<"-1";
			return 0;
		}
	}
	for(i=0;i<ans.size();i++){
		cout<<ans[i].first<<' '<<ans[i].second<<'\n';
	}
}