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>

int mymax(int x, int y){
        if(x>y){
                return x;
        }else{
                return y;
        }
}
int mymin(int x, int y){
        if(x>y){
                return y;
        }else{
                return x;
        }
}

int main(){
        int N;
        scanf("%d", &N);
        int a[50]={};
        for(int i=0;i<N;i++){
                scanf("%d", &a[i]);
        }
        int max=0,min=0,absolute=0,keep=0;
        for(int i=0;i<N-1;i++){
                if(max<mymax(a[i],a[i+1])){
                        max = mymax(a[i],a[i+1]);
                        keep=i+1;
                }
                if(min>mymin(a[i],a[i+1])){
                        min = mymin(a[i],a[i+1]);
                        keep=i+1;
                }
                if(abs(max)>abs(min)){
                        absolute = max;
                }else{
                        absolute = min;
                }
        }
        printf("%d\n", 2*N-1);
        for(int i=0;i<N;i++){
                a[i]+=absolute;
                printf("%d %d\n", keep+1, i+1);
        }
        int perma=0;
        perma = a[0];
        for(int i=0;i<N-1;i++){
                if(perma>=0){
                        a[i+1]+=a[i];
                        printf("%d %d\n", i+2, i+1);
                }else{
                        a[N-i-2]+=a[N-i-1];
                        printf("%d %d\n", N-i-1,N-i);
                }
        }
        return 0;
}