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 C {

	public static void main(String[] args)  throws Exception{
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));

		int n = Integer.parseInt(f.readLine());
		int[] p = new int[n];
		int count = 0;

		StringTokenizer st = new StringTokenizer(f.readLine());

		for(int i = 0;i<n;i++){
			p[i] = Integer.parseInt(st.nextToken()) - 1;
		}
		st = new StringTokenizer(f.readLine());
		for(int i = 0;i<n;i++){
			if(Integer.parseInt(st.nextToken()) == 1) count ++;
		}
		int ans = 0;
		boolean[] visited = new boolean[n];
		Arrays.fill(visited, false);
		for(int i = 0;i<n;i++){
			if(visited[i]) continue;
			ans ++;
			Queue<Integer> q = new LinkedList<Integer>();
			q.add(i);
			visited[i] = true;
			while(!q.isEmpty()){
				int a = q.poll();
				if(!visited[p[a]]){
					q.add(p[a]);
					visited[p[a]] = true;
				}
			}
		}
		
		System.out.println((ans == 1 ?0:ans) + (count % 2 == 0? 1 : 0));



	}

}
