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[] arr = new int[n];
      int min = Integer.MAX_VALUE;
      int max = Integer.MIN_VALUE;
      int minIdx = 0, maxIdx = 0;

      for(int i = 0; i < n; i++) {
        arr[i] = sc.nextInt();
        if(arr[i] < min) {
          min = arr[i];
          minIdx = i;
        }
        if(arr[i] > max) {
          max = arr[i];
          maxIdx = i;
        }
      }
	  System.out.println(2*n-1);
      if(min < 0 && Math.abs(min) > Math.abs(max)) { // neg
        for(int i = 0; i < n; i++) {
          arr[i] += min;
          System.out.println((minIdx+1) + " " + (i+1));
        }
        for(int i = n-1; i >= 1; i--) {
          arr[i-1] += arr[i];
          System.out.println((i+1) + " " + i);
        }
      } else {
        for(int i = 0; i < n; i++) {
          arr[i] += max;
          System.out.println((maxIdx+1) + " " + (i+1));
        }
        for(int i = 1; i < n; i++) {
          arr[i] += arr[i-1];
          System.out.println(i + " " + (i+1));
        }
      }
  }
}