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<bits/stdc++.h>
#define ll long long
using namespace std;
int n,cnt,size[100005];
ll dis[100005];
struct node{
	ll d;
	int id;
}a[100005];
struct edge{
	int x,y;
}e[100005];

int cmp(node u,node v){
	return u.d<v.d;
}

int cmp1(edge u,edge v){
	return u.x<v.x||(u.x==v.x&&u.y<v.y);
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i].d),a[i].id=i;
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++) size[i]=1;
	for(int i=n;i>=2;i--){
		ll o=a[i].d+1ll*(2*size[a[i].id]-n);
		int l=1,r=i-1,mid,p=i;
		while(l<=r){
			mid=(l+r)/2;
			if(a[mid].d>=o) p=min(p,mid),r=mid-1;
			else l=mid+1;
		}
		if(p==i||a[p].d!=o){puts("-1");return 0;}
		size[a[p].id]+=size[a[i].id];
		e[++cnt]=(edge){a[p].id,a[i].id};
		if(e[cnt].x>e[cnt].y) swap(e[cnt].x,e[cnt].y);
		dis[a[p].id]+=dis[a[i].id]+1ll*size[a[i].id];
	}
	if(dis[a[1].id]!=a[1].d){puts("-1");return 0;}
	sort(e+1,e+cnt+1,cmp1);
	for(int i=1;i<=cnt;i++) printf("%d %d\n",e[i].x,e[i].y); 

	return 0; 
}