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 <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
ll dat[100005];
pair<ll,ll> p[100005];
ll sz[100005];
ll from[100005];
ll to[100005];
ll n;
int main() {
	cin>>n;
	for(ll i=0;i<n;i++){
		cin>>dat[i];
		p[i]=make_pair(dat[i],i);
	}
	sort(p,p+n);
	ll id=0;
	ll sum=0;
	for(int i=0;i<n;i++){
		sz[i]=1;
	}
	for(ll i=n-1;i>=1;i--){
		ll curid=p[i].second;
		ll tmp=p[i].first;
		ll nx=n-2*sz[curid];
		if(nx<=0){
			cout<<-1<<endl;
			return 0;
		}
		ll pos=lower_bound(p,p+n,make_pair(tmp-nx,0LL))-p;
		if(p[pos].first!=tmp-nx){
			cout<<-1<<endl;
			return 0;
		}
		from[id]=p[i].second;
		to[id++]=p[pos].second;
		sum+=sz[curid]*(n-sz[curid])*2;
		sz[p[pos].second]+=sz[curid];
	}
	ll ans=0;
	for(ll i=0;i<n;i++){
		ans+=dat[i];
	}
	if(sum!=ans) {
		cout<<-1<<endl;
		return 0;
	}
	for(ll i=0;i<id;i++){
		cout<<from[i]+1<<" "<<to[i]+1<<endl;
	}
	return 0;
}
