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>
using namespace std;
#define int long long
const int maxn=1e5+5;
inline int read(){
	char c=getchar();int t=0,f=1;
	while((!isdigit(c))&&(c!=EOF)){if(c=='-')f=-1;c=getchar();}
	while((isdigit(c))&&(c!=EOF)){t=(t<<3)+(t<<1)+(c^48);c=getchar();}
	return t*f;
}
int n; 
struct node{
	int id,d;
	node(int d1=0,int id1=0){d=d1,id=id1;}
	friend bool operator <(node a,node b){return a.d<b.d;}
}a[maxn];
int sz[maxn],u[maxn],v[maxn],dis,cnt;
signed main(){
	n=read();
	for(int i=1;i<=n;i++){
		a[i]=node(read(),i),sz[i]=1;
	}
	sort(a+1,a+1+n);
	for(int i=n;i>1;i--){
		int d=a[i].d-n+(sz[i]<<1);
		int p=lower_bound(a+1,a+1+n,node(d,0))-a;
		if(a[p].d!=d){puts("-1");return 0;}
		cnt++;
		u[cnt]=a[i].id;v[cnt]=a[p].id;
		sz[p]+=sz[i];
		dis+=sz[i];
	}
	if(a[1].d!=dis){
		puts("-1");return 0;
	}
	for(int i=1;i<=cnt;i++){
		printf("%d %d\n",u[i],v[i]);
	}
	return 0;
}
