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;
int n,sz[100005],fa[100005];
long long d[100005],res;
map<long long,int> id;
priority_queue<pair<long long,int> > pq;
vector<int> e[100005];
vector<pair<int,int> > ans;
void dfs(int x,int d){
	res+=d;
	for(int i=0;i<e[x].size();i++) dfs(e[x][i],d+1);
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		sz[i]=1;
		cin>>d[i];
		id[d[i]]=i;
		pq.push(make_pair(d[i],i));
	}
	while(pq.size()>1){
		int x=pq.top().second;pq.pop();
		id.erase(d[x]);
		if(!id.count(d[x]-(n-sz[x])+sz[x])){
			puts("-1");
			return 0;
		}
		fa[x]=id[d[x]-(n-sz[x])+sz[x]];
		sz[fa[x]]+=sz[x];
		e[fa[x]].push_back(x);
		ans.push_back(make_pair(fa[x],x));
	}
	int rt=pq.top().second;
	res=0;
	dfs(rt,0);
	if(res!=d[rt]){
		puts("-1"); 
	}
	else{
		for(int i=0;i<n-1;i++){
			printf("%d %d\n",ans[i].first,ans[i].second);
		}
	}
	return 0;
}