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
import java.util.Scanner;
import java.util.ArrayList;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int[] a=new int[n];
		int index=0;
		int max=0;
		for(int i=0; i<n; i++){
			a[i]=sc.nextInt();
			if(Math.abs(max)<Math.abs(a[i])){
				max=a[i];
				index=i;
			}
		}
		ArrayList<String> list=new ArrayList<>();
		if(max<0){
			for(int i=0; i<n; i++){
				if(a[i]>=0){
					a[i]+=max;
					list.add((index+1)+" "+(i+1));
				}
			}
		}else{
			for(int i=0; i<n; i++){
				if(a[i]<0){
					a[i]+=max;
					list.add((index+1)+" "+(i+1));
				}
			}
		}
		System.out.println(list.size()+n-1);
		for(int i=0; i<list.size(); i++){
			System.out.println(list.get(i));
		}
		if(max<0){
			for(int i=n-1; i>0; i--){
				System.out.println((i+1)+" "+i);
			}
		}else{
			for(int i=0; i<n-1; i++){
				System.out.println((i+1)+" "+(i+2));
			}
		}
	}
}
