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: Runtime Error)

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

class Main{
	public static void main(String[] args) {



		Scanner scanner = new Scanner(System.in);
		int N = scanner.nextInt();
		int[] data = new int[N];

		for(int i = 0;i < N;i++){
			data[i] = scanner.nextInt();
		}

		int[] copy = data.clone();


		Arrays.sort(copy);



		int[] plus = new int[N];
		int[] minus = new int[N];

		int p = 0;					//配列plusのインデックス
		int m = 0;					//配列mnusのインデックス

		int pnum = 0;				//配列plusの中の最大値のインデックス
		int mnum = 0;				//配列の中の最大値のインデックス

		for(int i = 0;i < N;i++){			//マイナスとプラスの値をそれぞれの配列にコピー
			if(copy[i] <= 0){
				minus[m] = copy[i];
				m++;
			}else{
				plus[p] = copy[i];
				p++;
			}
		}
		m--;
		p--;

		if(p == -1){
			p = 0;
		}
		if(m == -1){
			m = 0;
		}


		for(int i = 0;i < N;i++){				//絶対値の大きさが最大のとこのindexを見つける
			if( data[i] == plus[p] ){
				pnum = i;
			}else if(data[i] == minus[0]){
				mnum = i;

			}
		}


		int absplus = Math.abs(plus[p]);
		int absminus = Math.abs(minus[0]);

		int[] ans1 = new int[1000*N];
		int[] ans2 = new int[1000*N];
		int a1 = 0;
		int a2 = 0;

		if(absminus >= absplus){				//絶対値が一番大きいのがマイナスのやつ
			int key = minus[0];

			for(int i = 0;i < N;i++){
									//全部に最大値をマイナスして、全部を負の値にする
					data[i] -= absminus;
				
			}

			for(int i = 0;i < N-1;i++){

				if(data[i+1] < data[i]){

					data[i] += data[i+1];
					ans1[a1] = i+1;
					ans2[a2] = i;
					a1++;
					a2++;
					i = -1;

					//					for(int j = 0;j < N;j++){
					//						System.out.println(data[j]);		
					//					}
					//					System.out.println();

				}
			}



		}else{					//絶対値が一番大きいのがプラスのやつ

			int key = plus[p];
			
			for(int i = 0;i < N;i++){
								//全部に最大値をプラスして、全部を正の値にする
					data[i] += absplus;
				
			}
			for(int i = 0;i < N-1;i++)
				if(data[i+1] < data[i]){
					data[i+1] += data[i];
					ans1[a1] = i;
					ans2[a2] = i+1;
					a1++;
					a2++;
					i = -1;

				}
		}




		System.out.println(a1);

		for(int i = 0;i < a1;i++){
			System.out.println((ans1[i]+1) + " "+ (ans2[i]+1));
		}







	}




}

class Pair implements Comparable{
	int from;
	int end;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}