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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class C8VCFinals2017 {
	public static void main(String[] args) {
		FS scan = new FS(System.in);
		
		int N = scan.nextInt();
		int[] p = new int[N];
		for(int i=0;i<N;i++)p[i] = scan.nextInt()-1;
		int[] b = new int[N];
		for(int i=0;i<N;i++)b[i] = scan.nextInt();
		
		int count = 0;
		boolean[] seen = new boolean[N];
		for(int i=0;i<N;i++){
			if(seen[i])continue;
			int idx = i;
			count++;
			while(!seen[idx]){
				seen[idx] = true;
				idx = p[idx];
			}
		}
		int odd = 0;
		for(int i=0;i<N;i++)odd+=b[i];
		odd%=2;
		System.out.println((count==1?0:count)+(1^odd));
	}
	private static class FS {
		BufferedReader br;
		StringTokenizer st;
		public FS(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}
		String next(){
			while(st==null||!st.hasMoreElements()){
				try{st = new StringTokenizer(br.readLine());}
				catch(IOException e){e.printStackTrace();}
			}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() { return Double.parseDouble(next());}
	}
}
