Pavel cooks barbecue. There are n skewers, they lay on a brazier in a row, each on one of n positions. Pavel wants each skewer to be cooked some time in every of n positions in two directions: in the one it was directed originally and in the reversed direction.

Pavel has a plan: a permutation p and a sequence b1, b2, ..., bn, consisting of zeros and ones. Each second Pavel move skewer on position i to position pi, and if bi equals 1 then he reverses it. So he hope that every skewer will visit every position in both directions.

Unfortunately, not every pair of permutation p and sequence b suits Pavel. What is the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements? Note that after changing the permutation should remain a permutation as well.

There is no problem for Pavel, if some skewer visits some of the placements several times before he ends to cook. In other words, a permutation p and a sequence b suit him if there is an integer k (k ≥ 2n), so that after k seconds each skewer visits each of the 2n placements.

It can be shown that some suitable pair of permutation p and sequence b exists for any n.

Input

The first line contain the integer n (1 ≤ n ≤ 2·105) — the number of skewers.

The second line contains a sequence of integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation, according to which Pavel wants to move the skewers.

The third line contains a sequence b1, b2, ..., bn consisting of zeros and ones, according to which Pavel wants to reverse the skewers.

Output

Print single integer — the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements.

Examples

Input

4
4 3 2 1
0 1 1 1


Output

2


Input

3
2 3 1
0 0 0


Output

1

Note

In the first example Pavel can change the permutation to 4, 3, 1, 2.

In the second example Pavel can change any element of b to 1.
import java.util.*;

public class aa
{
  private static String getString(Scanner scanner)
  {
    String s = scanner.nextLine();
    if (0 == s.length())
      s = scanner.nextLine();
    return s;
  }

  private static int visit(int[] permutation, int n)
{
  int[] visites = new int[n+2];
  int j=0;
  for(int i=1; i <= n; i++)
  {
    if (0 == visites[i])
    {
      ++j;
      int q = i;
      for (int l=0; 0 == visites[q]; ++l)
       {
         visites[q] = 1;
         q = permutation[q];
          }
        }
  }
  if (j <= 1)
    return 0;
  return j;
}

  public static void main(String[] args)
  {
    try {
      StringBuilder sb = new StringBuilder();
      Scanner scanner = new Scanner(System.in);
      
int i=0, j=0, n = scanner.nextInt();
int[] permutation = new int[n+2];
int flags = 0;
  for(i=1; i <= n; i++)
  {
    permutation[i] = scanner.nextInt();
  }
  for(i=0; i < n; i++)
 {
   int a = scanner.nextInt();
   flags += (0 == a) ? 0 : 1;
  }
  if (0 == (flags & 1))
    ++j;

  j += visit(permutation, n);
      sb.append(j);
      System.out.print(sb.toString());
    }catch(Exception ex){
        ex.printStackTrace();      
    }
  }
}