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<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
#include<algorithm>
#define MAXN 100010
#define LL long long
using namespace std;
LL d[MAXN],dis[MAXN],siz[MAXN];
int n,cnt; 
map<LL,int> mp;
struct Edge{int u,v;}E[MAXN];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&d[i]);
	for(int i=1;i<=n;i++) mp[d[i]]=i;
	sort(d+1,d+n+1);
	for(int i=1;i<=n;i++) siz[i]=1;
	//d0=di+m-(n-m)=di+2m-n;
	for(int i=n;i>1;i--){ 
		int u=mp[d[i]];
	//	cout<<u<<' '<<d[i]<<endl;
		LL d0=d[i]+2*siz[u]-n;
		int v=mp[d0];
	//	cout<<"-----"<<v<<' '<<d0<<endl; 
		if(v){
			siz[v]+=siz[u];
			dis[v]+=(dis[u]+siz[u]);
			E[++cnt].u=min(u,v);
			E[cnt].v=max(u,v);
		}
		else{
			puts("-1");
			return 0;
		}
	}
	if(dis[mp[d[1]]]!=d[1]){
		puts("-1");
		return 0;
	}
	for(int i=1;i<=cnt;i++) printf("%d %d\n",E[i].u,E[i].v);
	return 0;
}