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>

long long int absyai(long long int a){return a>0?a:-a;}

int main(void){
    int N;
    scanf("%d",&N);
    long long int a[N];
    long long int maxabsyai=0;
    int po;
    for(int i=0;i<N;i++){
        scanf("%lld",&a[i]);
        if(absyai(a[i])>=maxabsyai){maxabsyai=absyai(a[i]);po=i;}
    }
    if(maxabsyai==0){printf("0\n");return 0;}
    int ans[2][200]={};
    int kosu=0;
    if(a[po]>0){
        for(int i=0;i<N;i++){
            a[i]+=a[po];
            ans[0][kosu]=po;
            ans[1][kosu]=i;
            kosu++;
            if(a[i]>a[po]&&po>i){
                a[po]+=a[i];
                ans[0][kosu]=i;
                ans[1][kosu]=po;
                kosu++;
            }
            else if(a[i]<a[po]&&po>i){
                a[po]+=a[po];
                ans[0][kosu]=po;
                ans[1][kosu]=po;
                kosu++;
            }
            else if(a[i]>a[po]&&po<i){
                po=i;
                ans[0][kosu]=po;
                ans[1][kosu]=po;
                kosu++;
            }
            else if(a[i]<a[po]&&po<i){
                a[i]+=a[po];
                ans[0][kosu]=po;
                ans[1][kosu]=i;
                kosu++;
                po=i;
            }
        }
    }
    else if(a[po]<0){
        for(int i=N-1;i>=0;i--){
            a[i]+=a[po];
            ans[0][kosu]=po;
            ans[1][kosu]=i;
            kosu++;
            if(a[i]<a[po]&&po<i){
                a[po]+=a[i];
                ans[0][kosu]=i;
                ans[1][kosu]=po;
                kosu++;
            }
            else if(a[i]>a[po]&&po<i){
                a[po]+=a[po];
                ans[0][kosu]=po;
                ans[1][kosu]=po;
                kosu++;
            }
            else if(a[i]<a[po]&&po>i){
                po=i;
                ans[0][kosu]=po;
                ans[1][kosu]=po;
                kosu++;
            }
            else if(a[i]>a[po]&&po>i){
                a[i]+=a[po];
                ans[0][kosu]=po;
                ans[1][kosu]=i;
                kosu++;
                po=i;
            }
        }
    }
    printf("%d\n",kosu);
    for(int i=0;i<kosu;i++){
        printf("%d %d\n",ans[0][i]+1,ans[1][i]+1);
    }
    //for(int i=0;i<N;i++)printf("%lld ",a[i]);
    return 0;
}