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.io.*;
import java.util.*;

public class Q3 {
  
  public static void main (String[] args) throws Exception {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    
    int n = Integer.parseInt(in.readLine());
    int[] skewers = new int[n];
    String [] next = in.readLine().split(" ");
    for (int i = 0; i < n; i++) skewers[i] = Integer.parseInt(next[i]) - 1;
    next = in.readLine().split(" ");
    
    boolean ok = false;
    for (String s : next) {
      if (s.equals("1")) ok = !ok;
    }
    int ans = 0;
    int groups = 0;
    int flips = 0;
    boolean[] vs = new boolean[n];
    for (int i = 0; i < n; i++)  {
      if (vs[i]) continue;
      
      int idx = skewers[i];
      while (idx != i) {
        vs[idx] = true;
        idx = skewers[idx];
      }
      
      groups++;
      vs[idx] = true;
    }
    
    if (!ok) ans++;
    if (groups > 1) ans += groups;
    
    System.out.println(ans);
    
    
    in.close();
  }
  
}