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: Runtime Error)

#include<stdio.h>
int main(){
  long n,i,max=0,min=0,ans1[200],ans2[200],count=0;
  long long int a[60];
  scanf("%ld", &n);
  for(i=0;i<n;i++){
    scanf("%lld",&a[i]);
    if(i==0){
      max=i;
      min=i;
    }
    if(a[max]<a[i])max=i;
    if(a[min]>a[i])min=i;
  }
  if(a[max]==a[min]){
    puts("0");
    return 0;
  }
  else if(abs(a[min])<=abs(a[max])){
    for(i=1;i<n;i++){
      while(a[i-1]>a[i]){
        a[i]+=a[max];
        ans1[count]=max+1;
        ans2[count]=i+1;
        count++;
      }
      max=i;
    }
  }
  else {
    for(i=n-2;i>=0;i--){
      while(a[i+1]<a[i]){
        a[i]+=a[min];
        ans1[count]=min+1;
        ans2[count]=i+1;
        count++;
      }
      min=i;
    }
  }
  printf("%ld\n",count);
  for(i=0;i<count;i++) printf("%ld %ld\n",ans1[i],ans2[i]);
  return 0;
}