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>
int main(void){
    int n,ans=0,max=0,min=0,maxi,mini;
    scanf("%d",&n);
    int a[n],x[n],y[n];
    for (int i=0; i<n; i++){
        scanf("%d",&a[i]);
        if (max<a[i]){
            max=a[i];
            maxi=i;
        }
        if (min>a[i]){
            min=a[i];
            mini=i;
        }
    }
    if (max==0 && min==0){
    } else if (max>min*-1){
        for (int i=0; i<2*n; i++){
            int did=0;
            for (int j=1; j<n; j++){
                if (a[j]<a[j-1]){
                    a[j]+=max;
                    x[ans]=maxi+1;
                    y[ans]=j+1;
                    ans++;
                    did=1;
                    break;
                }
            }
            if (did==0){
                break;
            }
        }
    } else {
        for (int i=0; i<2*n; i++){
            int did=0;
            for (int j=1; j<n; j++){
                if (a[j]<a[j-1]){
                    a[j-1]+=min;
                    x[ans]=mini+1;
                    y[ans]=j;
                    ans++;
                    did=1;
                    break;
                }
            }
            if (did==0){
                break;
            }
        }
    }
    printf("%d\n",ans);
    if (ans!=0){
        for (int i=0; i<ans; i++){
            printf("%d %d\n",x[i],y[i]);
        }
    }
    return 0;
}