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.
/*
Author - Ayub Subhaniya
Institute - DAI-ICT
*/
import java.io.*;
import java.util.*;
import java.math.*;
public class Venture8C {
	static InputReader in=new InputReader(System.in);
	static OutputWriter out=new OutputWriter(System.out);
	
	public static void main(String[] args) {
		int n=in.nextInt();
		Graph g=new Graph(n);
		for (int i=0;i<n;i++)
			g.addEdge(i, in.nextInt()-1);
		int b[]=new int[n];
		int cnt=0;
		for (int i=0;i<n;i++)
		{
			b[i]=in.nextInt();
			if (b[i]==1)
				cnt++;
		}
		out.println((cnt+1)%2+g.DFS());
		out.close();
	}
	
	
	
}
class Graph {
	LinkedList<Integer> adj[];
	int V;
	boolean visited[];

	Graph(int v) {
		V = v;
		adj = new LinkedList[v];
		visited = new boolean[v];
		for (int i = 0; i < v; i++) {
			adj[i] = new LinkedList<Integer>();
		}
	}

	void addEdge(int v, int u) {
		adj[v].add(u);
	}


	void DFSUtil(int v) {
		Iterator<Integer> it = adj[v].iterator();

		while (it.hasNext()) {
			int n = it.next();

			if (!visited[n]) {
				visited[n] = true;
				DFSUtil(n);
			}
		}
	}

	int DFS() {
		int comp=0;
		for (int i = 0; i < V; i++) {
			if (!visited[i]) {
				visited[i] = true;
				comp++;
				DFSUtil(i);
			}
		}
		return comp==1? 0 :comp;
	}

	

}

class InputReader {
	public BufferedReader reader;
	public StringTokenizer tokenizer;

	public InputReader(InputStream inputstream) {
		reader = new BufferedReader(new InputStreamReader(inputstream));
		tokenizer = null;
	}

	public String nextLine() {
		String fullLine = null;
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				fullLine = reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			return fullLine;
		}
		return fullLine;
	}

	public String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public long nextLong() {
		return Long.parseLong(next());
	}
	
	public int[] nextIntArray(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		return a;
	}
	
	public long[] nextLongArray(int n) {
		long a[] = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextLong();
		}
		return a;
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}
}

class OutputWriter {
	private final PrintWriter writer;

	public OutputWriter(OutputStream outputStream) {
		writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
	}

	public OutputWriter(Writer writer) {
		this.writer = new PrintWriter(writer);
	}

	public void print(Object... objects) {
		for (int i = 0; i < objects.length; i++) {
			if (i != 0)
				writer.print(' ');
			writer.print(objects[i]);
		}
	}

	public void println(Object... objects) {
		print(objects);
		writer.println();
	}

	public void close() {
		writer.close();
	}

	public void flush() {
		writer.flush();
	}
}