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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc=new Scanner(System.in);
		int n = sc.nextInt();
		long[] a=new long[n];
		long maxAbs=0;
		int argMaxAbs=0;
		for (int i=0;i<n;i++) {
			a[i]=sc.nextInt();
			if(Math.abs(a[i])>maxAbs) {
				maxAbs=Math.abs(a[i]);
				argMaxAbs=i;
			}
		}
		int count=0;
		int[] first=new int[2*n];
		int[] second=new int[2*n];
		if(a[argMaxAbs]>0) {
			for (int i=0;i<n-1;i++) {
				for(int j=0;j<2;j++) {
					first[count]=argMaxAbs+1;
					second[count]=i+2;
					count++;
					a[i+1]+=a[argMaxAbs];

				}
				
				if(Math.abs(a[i+1])>maxAbs) {
					maxAbs=Math.abs(a[i+1]);
					argMaxAbs=i+1;
				}		
			}
		}
		else if (a[argMaxAbs]<0){
			for (int i=n-1;i>0;i--) {
				for(int j=0;j<2;j++) {
					first[count]=argMaxAbs+1;
					second[count]=i;
					count++;
					a[i-1]+=a[argMaxAbs];

				}
				
				if(Math.abs(a[i-1])>maxAbs) {
					maxAbs=Math.abs(a[i-1]);
					argMaxAbs=i-1;
				}
			}
		}
		System.out.println(count);
		for(int i=0;i<count;i++) {
			System.out.println(first[i]+" "+second[i]);
		}
		
	}

}