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 <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=1e5+5;
int n,siz[N];
pair <int,int> Edge[N];
struct Node {
	ll dis;int id;
	bool operator < (const Node&b) const {return dis<b.dis;}
} A[N];

template <typename _Tp> inline void IN(_Tp&x) {
	char ch;bool flag=0;x=0;
	while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	if(flag) x=-x;
}

int main() {
	IN(n);
	for(int i=1;i<=n;++i) IN(A[i].dis),A[i].id=i;
	for(int i=1;i<=n;++i) siz[i]=1;
	sort(A+1,A+1+n);
	ll res=0;int cnt=0;
	for(int i=n;i>=2;--i) {
		ll dis=A[i].dis+(siz[i]<<1)-n;
		int pos=lower_bound(A+1,A+1+n,(Node){dis,0})-A;
		if(dis!=A[pos].dis) return puts("-1"),0;
		siz[pos]+=siz[i],Edge[++cnt]=make_pair(A[i].id,A[pos].id),res+=siz[i];
	}
	if(res!=A[1].dis) return puts("-1"),0;
	for(int i=1;i<=cnt;++i) printf("%d %d\n",Edge[i].first,Edge[i].second);
	return 0;
}