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)

//package work1;
import java.io.*;
import java.util.*;
import java.math.*;
public class Main
{
	public static void main(String[] args)
	{
		Scanner cin=new Scanner(System.in);
		int n=cin.nextInt();
		BigInteger []a=new BigInteger[n+1];
		int []x=new int[n*4];
		int []y=new int[n*4];
		int cnt=0;
		for(int i=1;i<=n;++i)
		{
			a[i]=cin.nextBigInteger();
			if(i>1)
			{
				if(a[i].compareTo(a[i-1])<0)
				{
					if(a[i].compareTo(BigInteger.ZERO)>0) 
					{
						x[cnt]=i-1; 
						y[cnt++]=i;
						a[i].add(a[i-1]);
					}
					else if(a[i].equals(BigInteger.ZERO))
					{
						x[cnt]=i-1;
						y[cnt++]=i;
					}
					else if(a[i-1].compareTo(BigInteger.ZERO)<=0)
					{
						x[cnt]=i;
						y[cnt++]=i-1;
					}
					else if(a[i].abs().compareTo(a[i-1].abs())>=0) 
					{
						x[cnt]=i; y[cnt++]=i-1;
						x[cnt]=i; y[cnt++]=i-1;
					}
					else 
					{
						x[cnt]=i-1; y[cnt++]=i;
						x[cnt]=i-1; y[cnt++]=i;
						a[i].add(a[i-1].multiply(BigInteger.valueOf(2)));
					}
				}
			}
		}
		System.out.println(cnt);
		for(int i=0;i<cnt;++i) System.out.println(x[i]+" "+y[i]);
	}
}