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>
#define int long long
using namespace std;
const int N=5000005;
struct ab{int f,t;}t[N];
int a[N],size[N],id[N],hh[N],h,tot;
map<int,int>mp;
void make(int x,int y){t[++h].f=hh[x],t[h].t=y,hh[x]=h;}
int cmp(int x,int y){return a[x]>a[y];}
void dfs(int x,int y,int z)
{
	tot+=z;
	for(int i=hh[x];i;i=t[i].f)
	{
		int j=t[i].t;
		if(j==y) continue;
		dfs(j,x,z+1);
	}
}
signed main()
{
	int n;
	scanf("%lld",&n);
	for(int i=1;i<=n;i++) scanf("%lld",a+i),mp[a[i]]=i,size[i]=1,id[i]=i;
	sort(id+1,id+1+n,cmp);
	for(int i=1;i<n;i++)
	{
		int x=id[i],y=a[x]-(n-size[x])+size[x];
		if(!mp[y]||y>=a[x]) return printf("-1\n"),0;
		y=mp[y];
		make(x,y),make(y,x);
		size[y]+=size[x];
	}
	dfs(id[n],0,0);
	if(tot!=a[id[n]]) return printf("-1\n"),0;
	for(int i=1;i<=n;i++)
	{
		for(int j=hh[i];j;j=t[j].f)
		{
			if(t[j].t>i) printf("%lld %lld\n",i,t[j].t);
		}
	}
	return 0;
}