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>

int main() {
    int N;
    scanf("%d", &N);
    long int a[100];
    int x[200], cnt=0; //x,yの値を入れていく
    for (int i=0; i < N; i++) { scanf("%ld", a+i); }
    for (int i=0; i < N -1; i++, cnt+=2) {
        if (a[i] > a[i+1]) {
            if (a[i] >= 0 && a[i+1] >= 0 ) {
                a[i+1] += a[i];
                x[cnt] = i+1;
                x[cnt+1] = i+2;
                //printf("%d %d", i+1, i+2); x,yがそれぞれi+1,i+2
            }else if (a[i] >= 0 && a[i+1] < 0) {
                a[i+1] += a[i];
                x[cnt] = i+1;
                x[cnt+1] = i+2;
                //printf("%d %d", i+1, i+2); x,yがそれぞれi+1,i+2
            }else if (a[i] < 0 && a[i+1] >= 0 ) {
                a[i+1] += a[i];
                x[cnt] = i+1;
                x[cnt+1] = i+2;
                //printf("%d %d", i+1, i+2); x,yがそれぞれi+1,i+2
            }else {
                a[i] += a[i+1];
                x[cnt] = i+2;
                x[cnt+1] = i+1;
                //printf("%d %d", i+2, i+1); x,yがそえぞれあi+2,i+1
            }
            i = 0; //iを0にして先頭から大小比較
        }else {
            cnt -= 2;
        }
    }
    printf("%d\n", cnt/2);
    for (int i=0; i < cnt; i+=2) {
        if (x[i] != 0) {
            printf("%d %d\n", x[i], x[i+1]);
        }else {
            continue;
        }
    }
    return 0;
}