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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

class Main {
    private static StringBuilder ans;
    private static int ops = 0;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        ans = new StringBuilder();
        int n = scanner.nextInt();
        long[] a = new long[n];
        for (int i = 0; i < a.length; i++) a[i] = scanner.nextInt();
        int[] endPoints = new int[]{0, a.length};
        while (endPoints.length > 0){
            endPoints = findAndResolve(a, endPoints[0], endPoints[1]);
        }
        System.out.println(ops);
        System.out.println(Arrays.toString(a));
        if (ans.length() > 1)ans.deleteCharAt(ans.length() - 1);
        System.out.println(ans.toString());
    }

    private static int[] findAndResolve(long[] a, int from, int upto){
        if (from == upto || upto < from) return new int[]{};
        int maxIndex = 0;
        long maxAbs = 0;
        long max = 0;

        for (int i = from; i < upto; i++) {
            if (Math.abs(a[i]) > maxAbs){
                maxIndex = i;
                max = a[i];
                maxAbs = Math.abs(a[i]);
            }
        }

        if (max > 0){
            for (int i = maxIndex; i < upto; i++) {
                if (a[i] < max){
                    a[i] += max;
                    ops ++;
                    ans.append(maxIndex + 1).append(" ").append(i + 1).append("\n");
                    if (a[i] < max){
                        a[i] += max;
                        ops++;
                        ans.append(maxIndex + 1).append(" ").append(i + 1).append("\n");
                    }
                }
            }
            return new int[]{from, maxIndex};
        }
        else if(max < 0){
            for (int i = maxIndex; i >= from; i--) {
                if (a[i] > max){
                    a[i] += max;
                    ops++;
                    ans.append(maxIndex + 1).append(" ").append(i + 1).append("\n");
                    if (a[i] > max){
                        a[i] += max;
                        ops++;
                        ans.append(maxIndex + 1).append(" ").append(i + 1).append("\n");
                    }
                }
            }
            return new int[]{maxIndex + 1, upto};
        }

        return new int[]{};
    }
}