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 <string.h>
#include <math.h>
int a[50], op[100][2], c=0;

void ope(int i, int j){
  //printf("mark: %d %d\n",i,j);
  op[c][0]=i;  op[c][1]=j; c++;
  a[j]+=a[i];
  return;
}

int main(void){
  int n, i;
  scanf("%d",&n);
  for(i=0;i<n;i++) scanf("%d",&a[i]);
  for(i=0;i<n-1;i++){
    if(a[i]>a[i+1]){
      if(a[i]*a[i+1]>0) ope(i+2,i+1);
      else if(a[i]*a[i+1]<0){ope(i+1,i+2); ope(i+1,i+2);}
      else{ope(i+2,i+1); ope(i+2,i+1);}
    }
  }
  printf("%d\n",c);
  for(i=0;i<c;i++) printf("%d %d\n",op[i][0],op[i][1]);
  return 0;
}