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 CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

int main()
{
  int n;
  int a[50];
  
  int max_abs_pos = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (i && abs(a[i]) > abs(a[max_abs_pos])) {
      max_abs_pos = i;
    }
  }
  
  printf("%d\n", 2 * (n - 1));
  for (int i = 0; i < n; i++) {
    if (i != max_abs_pos) printf("%d %d\n", max_abs_pos + 1, i + 1);
  }
  
  for (int i = 0; i < n - 1; i++) {
    if (a[max_abs_pos] >= 0) {
      printf("%d %d\n", i + 1, i + 2);
    }
    else {
      printf("%d %d\n", n - i, n - i - 1);
    }
  }
  
  return 0;
}