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
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n;
signed main(){
	
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin>>n;
	vector<int>d(n);
	vector<int>w(n,1);
	map<int,int>mm;
	for(int i=0;i<n;i++){
		cin>>d[i];
		mm[d[i]]=i;
	}
	vector<pii>res;
	for(int iit=0;iit<n-1;iit++){
		auto it=mm.rbegin();
		int i=it->S;
		int x=d[i]-(n-w[i])+w[i];
		if(!mm.count(x)||mm[x]==i)return puts("-1"),0;
		int j=mm[x];
		res.pb(mp(i,j));
		w[j]+=w[i];
		mm.erase(--mm.end());
	}
	vector<int>v(n,-1);
	vector<vector<int> >G(n);
	v[0]=0;
	for(auto&p:res){
		G[p.F].pb(p.S);
		G[p.S].pb(p.F);
	}
	vector<int>qq(1,0);
	for(int b=0;b<(int)(qq.size());b++)
		for(int j:G[qq[b]])
			if(v[j]==-1){
				qq.pb(j);
				v[j]=v[qq[b]]+1;
			}
	if(accumulate(v.begin(),v.end(),0LL)!=d[0])return puts("-1"),0;
	for(auto&p:res)cout<<p.F+1<<' '<<p.S+1<<endl;
	return 0;
}
