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: Accepted)

#include <stdio.h>
#include <stdlib.h>

#define N	100000

long long ss[N];

int compare(const void *a, const void *b) {
	int i = *(int *) a;
	int j = *(int *) b;

	return ss[i] < ss[j] ? -1 : 1;
}

int search(int *ii, int n, long long s) {
	int lower = -1, upper = n;

	while (upper - lower > 1) {
		int h = (lower + upper) / 2;

		if (ss[ii[h]] <= s)
			lower = h;
		else
			upper = h;
	}
	return lower == -1 || ss[ii[lower]] != s ? -1 : ii[lower];
}

int main() {
	static int ii[N], sz[N], pp[N], dd[N];
	int n, i;
	long long sum;

	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%lld", &ss[i]);
		ii[i] = i;
	}
	qsort(ii, n, sizeof *ii, compare);
	for (i = 0; i < n; i++)
		sz[i] = 1;
	for (i = n - 1; i > 0; i--) {
		int p = search(ii, n, ss[ii[i]] - (n - sz[ii[i]] * 2));

		if (sz[ii[i]] * 2 >= n || p == -1) {
			printf("-1\n");
			return 0;
		}
		pp[ii[i]] = p, sz[p] += sz[ii[i]];
	}
	sum = 0;
	for (i = 1; i < n; i++)
		sum += dd[ii[i]] = dd[pp[ii[i]]] + 1;
	if (ss[ii[0]] != sum) {
		printf("-1\n");
		return 0;
	}
	for (i = 1; i < n; i++)
		printf("%d %d\n", ii[i] + 1, pp[ii[i]] + 1);
	return 0;
}