Problem: 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.

Examples

Input

3
-2 5 -1


Output

2
2 3
3 3


Input

2
-1 -3


Output

1
2 1


Input

5
0 0 0 0 0


Output

0

Solution (in JAVA):

Solution:
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[2*N];
		int[] ans2 = new int[2*N];
		int a1 = 0;
		int a2 = 0;

		if(Math.abs(minus[0]) > plus[p]){				//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] += key;
				ans1[a1] = mnum;
				ans2[a2] = i;
				a1++;
				a2++;

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


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

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

				//					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] += key;
				ans1[a1] = pnum;
				ans2[a2] = i;
				a1++;
				a2++;

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

			for(int i = 0;i < N-1;i++){
				data[i+1] += data[i];
				ans1[a1] = i;
				ans2[a2] = i+1;
				a1++;
				a2++;


			}
		}




		System.out.println(a1);

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

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




	}




}

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

		return end - otherpair.end;
	}
}