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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define re register
#define ll long long
inline int gi()
{
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=100010;
typedef pair<ll,int> pli;typedef pair<int,int> pii;
#define mp make_pair
pli p[N];int n,siz[N];
vector<pii>ans;
int main()
{
	n=gi();
	for(int i=1;i<=n;i++)scanf("%lld",&p[i].first),p[i].second=i,siz[i]=1;
	sort(p+1,p+n+1);ll dis=0;
	for(int i=n;i>1;i--)
	{
		ll nd=p[i].first-(n-siz[p[i].second])+siz[p[i].second];
		int pos=lower_bound(p+1,p+n+1,mp(nd,0))-p;
		if(p[pos].first!=nd)return puts("-1"),0;
		ans.push_back(mp(p[pos].second,p[i].second));
		siz[p[pos].second]+=siz[p[i].second];dis+=siz[p[i].second];
	}
	if(dis!=p[1].first)return puts("-1"),0;
	for(pii now:ans)printf("%d %d\n",now.first,now.second);
	return 0;
}