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: Wrong Answer)

#include <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>

using namespace std;

const int INF = 1e9;

vector < pair <int, int> > v;

int n, sz[1000000], r[1000000];

bool ans = true;

vector <int> g[1000000];

int dfs (int x, int p)
{
	int sum = 1;

	for (int to : g[x])
		if (to != p)
			sum += dfs (to, x);

	if (r[x] != sum) ans = false;
}

int main () 
{
    cin >> n;
    
    for (int i = 0; i < n; i++)
    {
		int d;
		scanf ("%d", &d);
		
		v.emplace_back (d, i);

		r[i] = d;
    }
    
    sort (v.begin (), v.end ());
    
    for (int i = n - 1; i >= 0; i--)
    {
		int x = v[i].second, d = v[i].first;
		
		v.pop_back ();
		
		sz[x] = 1;
		
		for (int to : g[x])
		    sz[x] += sz[to];
		
		if (!i) break;
		
		int k = lower_bound (v.begin (), v.end (), make_pair (d + 2 * sz[x] - n, -1)) - v.begin ();
		
		if (k == v.size () || v[k].first != d + 2 * sz[x] - n)
		{
		    cout << -1;
		    return 0;
		}
		
		g[v[k].second].push_back (x);
    }
    
    dfs (0, -1);
    
    if (!ans)
    {
		cout << -1;
		return 0;
    }
    
    for (int i = 0; i < n; i++)
		for (int to : g[i])
		{
		    printf ("%d %d\n", i + 1, to + 1);
		}
}