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.*;

class Main 
{
  	public static void main(String[] args) 
  	{
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();

		int arr[] = new int[n];
		for(int i = 0; i < n; i++)
			arr[i] = sc.nextInt();

		boolean ok;
		LinkedList<String> solve = new LinkedList<String>();
		
		do
		{
			ok = true;

			for(int i = 0; i < n - 1; i++)
			{
				if(arr[i] > arr[i + 1])
				{
					ok = false;

					if(arr[i] < 0 && arr[i + 1] < 0)
					{
						arr[i] += arr[i + 1];
						solve.add((i + 2) + " " + (i + 1));
					}
					else
					{
						arr[i + 1] += arr[i];
						solve.add((i + 1) + " " + (i + 2));
					}
				}
			}
		}while(!ok);

		System.out.println(solve.size());
		if(solve.size() != 0)
			for(String s : solve)
				System.out.println(s);
  	}
}