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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

vector<int> v[100010];
int n,id[100010],l,r,mid,ans,siz[100010];
long long dis[100010],t;

bool cmp(int x,int y) {return dis[x]>dis[y];}

long long dfs(int x)
{
	long long ans=0;
	for (int i=0,sz=v[x].size(); i<sz; i++) ans+=dfs(v[x][i])+siz[v[x][i]];
	return ans;
}

int main()
{
	scanf("%d",&n);
	for (int i=1; i<=n; i++) scanf("%lld",&dis[i]),id[i]=i;
	sort(id+1,id+1+n,cmp);
	for (int i=1; i<=n; i++) siz[i]=1;
	for (int I=1,i; I<n; I++)
	{
		i=id[I],t=dis[i]-n+2*siz[i],l=I+1,r=n,ans=-1;
		while (l<=r) {mid=(l+r)>>1; if (dis[id[mid]]<=t) r=mid-1,ans=mid; else l=mid+1;}
		if (ans==-1||dis[id[ans]]!=t) return puts("-1"),0;
		v[id[ans]].push_back(i),siz[id[ans]]+=siz[i];
	}
	if (dfs(id[n])!=dis[id[n]]) return puts("-1"),0;
	for (int i=1; i<=n; i++)
		for (int j=0,sz=v[i].size(); j<sz; j++) printf("%d %d\n",i,v[i][j]);
	return 0;
}