Problem:
Score : 600 points

Problem Statement

Snuke has an integer sequence, a, of length N. The i-th element of a (1-indexed) is a_{i}.

He can perform the following operation any number of times:

Operation: Choose integers x and y between 1 and N (inclusive), and add a_x to a_y.

He would like to perform this operation between 0 and 2N times (inclusive) so that a satisfies the condition below. Show one such sequence of operations.
It can be proved that such a sequence of operations always exists under the constraints in this problem.

Condition: a_1 \leq a_2 \leq ... \leq a_{N}

Constraints

2 \leq N \leq 50

-10^{6} \leq a_i \leq 10^{6}

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Let m be the number of operations in your solution. In the first line, print m.
In the i-th of the subsequent m lines, print the numbers x and y chosen in the i-th operation, with a space in between.
The output will be considered correct if m is between 0 and 2N (inclusive) and a satisfies the condition after the m operations.

Sample Input 1

3
-2 5 -1

Sample Output 1

2
2 3
3 3

After the first operation, a = (-2,5,4).

After the second operation, a = (-2,5,8), and the condition is now satisfied.

Sample Input 2

2
-1 -3

Sample Output 2

1
2 1

After the first operation, a = (-4,-3) and the condition is now satisfied.

Sample Input 3

5
0 0 0 0 0

Sample Output 3

0

The condition is satisfied already in the beginning.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <math.h>
#include <limits.h>
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) > (b) ? (b) : (a))
#define abs(x) ((x) > 0 ? (x) : -(x))
#define MOD 1000000007 //10^9 + 7
#define endl printf("\n")
typedef long long ll;
#define MAX_N (1 << 17)

int
main(int argc, const char *argv[])
{
  int n, a[100];
  scanf("%d", &n);
  int min = 0, max = 0, p = 0, m = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] < 0) m++;
    if (a[i] > 0) p++;
    if (a[min] > a[i]) min = i;
    if (a[max] < a[i]) max = i;
  }

  if (a[min] >= 0) {
    printf("%d\n", n - 1);
    for (int i = 1; i < n; i++) printf("%d %d\n", i, i + 1);
    return 0;
  }

  else if (a[max] <= 0) {
    printf("%d\n", n - 1);
    for (int i = n - 1; i >= 1; i--) printf("%d %d\n", i + 1, i);
    return 0;
  }

  else {
    if (max >= -min) {
      printf("%d\n", m + n - 1);
      for (int i = 0; i < n; i++) {
	if (a[i] < 0) {
	  printf("%d %d\n", max + 1, i + 1);
	}
      }
      for (int i = 1; i < n; i++) printf("%d %d\n", i, i + 1);
      return 0;
    } else {
      printf("%d\n", p + n - 1);
      for (int i = 0; i < n; i++) {
	if (a[i] > 0) {
	  printf("%d %d\n", min + 1, i + 1);
	}
      }
      for (int i = n - 1; i >= 0; i--) printf("%d %d\n", i + 1, i);
      return 0;
    }
  }
}