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.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		
		int max = 1;
		int a[] = new int[n+1];
		
		for(int i=1;i<=n;i++){
			a[i] = sc.nextInt();
			if(Math.abs(a[i]) > Math.abs(a[max])){
				max = i;
			}
		}
		
		
		if(a[max] == 0){
			System.out.println(0);
		}
		else{
			int cnt = 0;
			int ans[][] = new int[2*n][2];
			
			for(int i=1;i<=n;i++){
				if(i!=max){
					ans[cnt][0] = max;
					ans[cnt][1] = i;
					a[i] += a[max];
					cnt++;
				}
			}
			
			if(a[max] > 0){
				for(int i=1;i<n;i++){
					if(a[i] > a[i+1]){
						ans[cnt][0] = i;
						ans[cnt][1] = i+1;
						a[i+1] += a[i];
						cnt++;
					}
				}
			}
			else{
				for(int i=n;i>1;i--){
					if(a[i-1] > a[i]){
						ans[cnt][0] = i;
						ans[cnt][1] = i-1;
						a[i-1] += a[i];
						cnt++;
					}
				}
			}
			
			System.out.println(cnt);
			for(int i=0;i<cnt;i++){
				System.out.println(ans[i][0] + " " +ans[i][1]);
			}
		}
	}

}