Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<unordered_map>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int N=1e5+10;
ll d[N];
unordered_map<ll,vector<int> > p;
ll cnt[N],sum[N],n,r[N];
vector<pair<int,int> > w;
bool comp(int a,int b){
	return d[a]>d[b];
}
int main(){
	cin>>n;
	for (int i=1;i<=n;i++)scanf("%lld",&d[i]),r[i]=i,p[d[i]].push_back(i),cnt[i]=1;
	sort(r+1,r+n+1,comp);
	for (int i=1;i<n;i++){
		ll x=r[i],s=d[x]-(n-2*cnt[x]),y;
		if (p[s].empty()||s>=d[x]){
			cout<<-1<<endl;
			return 0;
		}
		y=p[s].back();
		w.push_back(make_pair(x,y));
		cnt[y]+=cnt[x];
		sum[y]+=sum[x]+cnt[x];
	}
	if (sum[r[n]]!=d[r[n]]){
		cout<<-1<<endl;
		return 0;
	}
	for (int i=0;i<n-1;i++)printf("%d %d\n",w[i].first,w[i].second);
}