Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

7
10
15
13
18
11
14
19

Sample Output 1

1 2
1 3
1 5
3 4
5 6
6 7

The tree shown below satisfies the conditions.

Sample Input 2

2
1
2

Sample Output 2

-1

Sample Input 3

15
57
62
47
45
42
74
90
75
54
50
66
63
77
87
51

Sample Output 3

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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+3;
int n,siz[N];
bool b[N];
struct syzs{
	int val,num;
} a[N];
struct ans{
	int e1,e2;
	bool operator <(const ans &emm) const{
	    if(emm.e1==e1) return e2>emm.e2;
		return emm.e1<e1;
	}
};
bool cmp(syzs c,syzs b){
	return c.val>b.val;
}
int find(int k){
	int chec=a[k].val+2*siz[k]-n;
	for(int i=k+1;i<=n;i++){
		if(a[i].val==chec&&i!=k) return i;
	}
	return -1;
}
int main(){
	priority_queue <ans> q;
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i].val);
		a[i].num=i;
		siz[i]=1;
	}
	sort(a+1,a+n+1,cmp);
	b[1]=1;
	for(int i=1;i<n;i++){
		int v=find(i);
		if(v==-1) puts("-1"),exit(0);
		siz[v]+=siz[i];
		q.push((ans) {min(a[v].num,a[i].num),max(a[v].num,a[i].num)});
	}
	while(!q.empty()){
		int v1=q.top().e1,v2=q.top().e2;
    	q.pop();
    	printf("%d %d\n",v1,v2);
	}
}