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.PrintWriter;
import java.util.*;

public class Main {
	public static final int MAXN=200000+10;
	public static int p[]=new int [MAXN];
	public static int b[]=new int [MAXN];
	public static boolean vis[]=new boolean [MAXN];
	public static void dfs(int u){
		if(vis[u]==true){
			return;
		}
		vis[u]=true;
		dfs(p[u]);
	}
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n=cin.nextInt();
		for(int i=1;i<=n;i++){
			p[i]=cin.nextInt();
		}
		int sumb=0;
		for(int i=1;i<=n;i++){
			b[i]=cin.nextInt();
			sumb+=b[i];
		}
		int ans=0;
		Arrays.fill(vis, false);
		for(int i=1;i<=n;i++){
			if(vis[i]==false){
				ans++;
				dfs(i);
			}
		}
		if(ans!=1){
			ans=ans+(sumb%2==1?0:1);
		}else{
			ans=(sumb%2==1?0:1);
		}
		out.println(ans);
		cin.close();
		out.flush();
	}
}