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 N 100005
#define ll long long
#define pii pair<ll,int>
#define mp make_pair
using namespace std;

ll n,size[N];
int e[N][2];
pii a[N];

inline ll read()
{
	ll ans=0,f=1;char c=getchar();
	while(c>'9'||c<'0') {if(c=='-') f=-1ll;c=getchar();}
	while(c>='0'&&c<='9') ans=(ans<<1)+(ans<<3)+(c^48),c=getchar();
	return ans*f;
}

int main()
{
	n=read();
	for(int i=1;i<=n;++i) a[i]=mp(read(),i),size[i]=1;
	sort(a+1,a+1+n);
	for(int i=n;i>1;i--)
	{
		ll cur=a[i].first-n+2*size[i];
		int l=1,r=i-1;
		while(l<r)
		{
			int mid=l+r>>1;
			if(a[mid].first<cur) l=mid+1;
			else r=mid;
		}
		if(a[l].first!=cur) return puts("-1"),0;
		e[i][0]=a[i].second;
		e[i][1]=a[l].second;
		size[l]+=size[i];
	}
	for(int i=2;i<=n;++i) a[1].first-=size[i];
	if(a[1].first) return puts("-1"),0;
	for(int i=2;i<=n;++i) printf("%d %d\n",e[i][0],e[i][1]);
}
