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 = 0;
		int maxIdx = 0;
		int min = Integer.MAX_VALUE;
		int minIdx = 0;
		for (int i = 1; i <= n; i++) {
		    int x = sc.nextInt();
		    if (max < x) {
		        max = x;
		        maxIdx = i;
		    }
		    if (min > x) {
		        min = x;
		        minIdx = i;
		    }
		}
		boolean reverse;
		StringBuilder sb = new StringBuilder();
		if (min >= 0) {
		    sb.append(n - 1).append("\n");
		    reverse = false;
		} else if (max <= 0) {
		    sb.append(n - 1).append("\n");
		    reverse = true;
		} else if (Math.abs(min) <= Math.abs(max)) {
		    sb.append(n * 2 - 1).append("\n");
		    for (int i = 1; i <= n; i++) {
		        sb.append(maxIdx).append(" ").append(i).append("\n");
		    }
		    reverse = false;
		} else {
		    sb.append(n * 2 - 1).append("\n");
		    for (int i = 1; i <= n; i++) {
		        sb.append(minIdx).append(" ").append(i).append("\n");
		    }
		    reverse = true;
		}
		if (reverse) {
		    for (int i = n - 1; i >= 1; i--) {
		        sb.append(i + 1).append(" ").append(i).append("\n");
		    }
		} else {
		    for (int i = 2; i <= n; i++) {
		        sb.append(i - 1).append(" ").append(i).append("\n");
		    }
		}
       System.out.print(sb);
   }
}