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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define X 1000001

int main(void){
    int N,*a,i,max=-X,min=X,maxi,mini,count=0,**out;
    scanf("%d",&N);
    a = (int*)malloc(sizeof(int)*N);
    out = (int**)malloc(sizeof(int*)*(2*N));
    for(i=0;i<2*N;i++)    out[i] = (int*)malloc(sizeof(int)*2);
    for(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(min<0&&max>0){
        if(max>=-min){
            for(i=0;i<N;i++){
                a[i] += max;
                out[count][0] = maxi + 1;
                out[count][1] = i + 1;
                count++;
            }
            max += max;
            min += max;
        }else{
            for(i=0;i<N;i++){
                a[i] += min;
                out[count][0] = mini + 1;
                out[count][1] = i + 1;
                count++;
            }
            max += min;
            min += min;
        }
    }
    if(min>=0){
        for(i=1;i<N;i++){
            if(a[i-1]>a[i]){
                a[i] += a[i-1];
                out[count][0] = i;
                out[count][1] = i + 1;
                count++;
            }
        }
    }else{
        for(i=N-2;i>=0;i--){
            if(a[i]>a[i+1]){
                a[i] += a[i+1];
                out[count][0] = i + 2;
                out[count][1] = i + 1;
                count++;
            }
        }
    }
    printf("%d\n",count);
    for(i=0;i<count;i++){
        printf("%d %d\n",out[i][0],out[i][1]);
    }
    return 0;
}