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: Accepted)

#include <stdio.h>
#include <stdlib.h>
//#include <string.h>
//#include <stdbool.h>
//#include <limits.h>
//#include <math.h>

int main() {
  int n;
  scanf("%d", &n);
  int a[n], min = 0, max = 0, mini, maxi, p, pi;
  for (int i=0; i<n; i++) {
    scanf("%d", a+i);
    if (a[i]<min) {
      min = a[i];
      mini = i;
    }
    if (a[i]>max) {
      max = a[i];
      maxi = i;
    }
  }
  if (abs(min)<abs(max)) {
    p = max;
    pi = maxi;
  } else {
    p = min;
    pi = mini;
  }
  max += p;
  min += p;
  printf("%d\n", 2*n-1);
  for (int i=0; i<n; i++) {
    a[i] += p;
    printf("%d %d\n", pi+1, i+1);
  }
  if (min >= 0) {
    for (int i=1; i<n; i++)
      printf("%d %d\n", i, i+1);
  } else {
    for (int i=1; i<n; i++)
      printf("%d %d\n", n-i+1, n-i);
  }
  return 0;
}