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.*;

public class Main {

	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] arr = new int[n];
		int max = 0;
		int maxIdx = 0;
		boolean isPlus = false;
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
			int x = Math.abs(arr[i]);
			if (x > max) {
				max = x;
				maxIdx = i + 1;
				isPlus = (arr[i] >= 0);
			}
		}
		if (maxIdx == 0) {
			System.out.println(0);
			return;
		}
		int count = 0;
		StringBuilder sb = new StringBuilder(); 
		for (int i = 1; i <= n; i++) {
			if (maxIdx != i){
				sb.append(maxIdx).append(" ").append(i).append("\n");
				count++;
			}
		}
		if (isPlus) {
			for (int i = 1; i <= n - 1; i++) {
				sb.append(i).append(" ").append(i + 1).append("\n");
				count++;
			}
		} else {
			for (int i = n; i >= 2; i--) {
				sb.append(i).append(" ").append(i - 1).append("\n");
				count++;
			}
		}
		System.out.println(count);
		System.out.print(sb);
	}
}
