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.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String []args){
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		Task solver = new Task();
		solver.solve(1,in,out);
		out.close();
	}
	static class Task{
		public void solve(int testNumber,Scanner in,PrintWriter out){
			int n = in.nextInt();
			int Max = 0;
			int Min = (int)(1e6 + 10);
			int posMax = 0;
			int posMin = 0;
			for(int i = 0 ; i < n ; i++){
				int x = in.nextInt();
				if(x > Max) {
					Max = x;
					posMax = i + 1;
				}
				if(x < Min){
					Min = x;
					posMin = i + 1;
				}
			}
			if(Math.abs(Max) >= Math.abs(Min)){
				out.println(2 * n);
				for(int i = 1 ; i <= 2 ; i++){
					out.println(posMax + " " + 1);
				}
				for(int i = 2 ; i <= n ; i++){
					for(int j = 1 ; j <= 2 ; j++){
						out.println((i - 1) + " " + i);
					}
				}
			} else {
				out.println(2 * n);
				for(int i = 1 ; i <= 2 ; i++){
					out.println(posMin + " " + n);
				}
				for(int i = n ; i >= 2 ; i--){
					for(int j = 1 ; j <= n ; j++){
						out.println(i + " " + (i - 1));
					}
				}
			}
		}
	}
}