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;

int n,d[200010],dp[200010],sz[200010];
map<int,int> mp;
vector<pair<int,int> > edges,ans;

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++){cin>>d[i];mp[d[i]]=i;edges.push_back(make_pair(d[i],i));}
	sort(edges.begin(),edges.end());
	reverse(edges.begin(),edges.end());
	for(int i=1;i<=n;i++){sz[i]=1;dp[i]=0;}
	for(int i=0;i<edges.size()-1;i++){
		if(sz[edges[i].second]*2>=n){
			cout<<-1<<endl;
			return 0;
		}
		int need=edges[i].first-(n-2*sz[edges[i].second]);
		if(!mp.count(need)){
			cout<<-1<<endl;
			return 0;
		}
		dp[mp[need]]+=dp[edges[i].second]+sz[edges[i].second];
		sz[mp[need]]+=sz[edges[i].second];
		ans.push_back(make_pair(mp[need],edges[i].second));
	}
	if(dp[edges[edges.size()-1].second]!=edges[edges.size()-1].first){
		cout<<-1<<endl;
		return 0;
	}
	for(int i=0;i<ans.size();i++){
		cout<<ans[i].first<<" "<<ans[i].second<<endl;
	}
	return 0;
}