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>
#include <string.h>
#include <math.h>
#include <stdlib.h>


int main() 
{
   int n,i,m=0,j=0;
    scanf("%d ",&n);
    int a[n],b[2*n][2];
    for(i=0;i<n;i++)
        scanf("%d ",&a[i]);
    for(i=1;i<n-1;i++)
    {
        if(a[i]>a[i+1])
        {
            while(a[i+1]<a[i])
        {
            m++;
            a[i+1]+=a[i];
            b[j][0]=i;
            b[j][1]=i+1; 
            j++;    
        }
        }
    }
    printf("%d\n",m);
    for(i=0;i<m;i++)
    {
        printf("%d %d\n",b[i][0]+1,b[i][1]+1);
    }


    return 0;
}