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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.InputMismatchException;


public class CF_C {
	static FastScanner sc = new FastScanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	
	public static void main(String[] args) {
		new CF_C().run();
		out.close();
	}
	
	public void run(){
		int n = sc.nextInt();
		DSU dsu = new DSU(n);
		for(int i = 0;i<n;i++){
			int tmp = sc.nextInt();
			dsu.union(i, tmp-1);
		}
		int one = 0;
		for(int i = 0;i<n;i++){
			int tmp = sc.nextInt();
			if(tmp == 1)
				one++;
		}
		if((one&1)!=0){
			if(dsu.getRoots().size()==1)
				out.println(0);
			else
				out.println(dsu.getRoots().size());
		}
		else{
			if(dsu.getRoots().size()==1)
				out.println(1);
			else
				out.println(dsu.getRoots().size()+1);
		}
	}

	static class FastScanner {
		private final InputStream stream;
	    private final byte[] buf = new byte[8192];
	    private int curChar, snumChars;

	    public FastScanner(InputStream st) {
	      this.stream = st;
	    }

	    //读一个字节
	    public int read() {
	      if (snumChars == -1)
	        throw new InputMismatchException();
	      if (curChar >= snumChars) {
	        curChar = 0;
	        try {
	          snumChars = stream.read(buf);
	        } catch (IOException e) {
	          throw new InputMismatchException();
	        }
	        if (snumChars <= 0)
	          return -1;
	      }
	      return buf[curChar++];
	    }

	    public int nextInt() {
	      int c = read();
	      while (isSpaceChar(c)) {
	        c = read();
	      }
	      int sgn = 1;
	      if (c == '-') {
	        sgn = -1;
	        c = read();
	      }
	      int res = 0;
	      do {
	        res *= 10;
	        res += c - '0';
	        c = read();
	      } while (!isSpaceChar(c));
	      return res * sgn;
	    }

	    public boolean isSpaceChar(int c) {
	      return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	    }
	}
}
class DSU {
	int[] p;

	public DSU(int n) {
		p = new int[n];
		clear();
	}

	public void clear() {
		for (int i = 0; i < p.length; i++) {
			p[i] = i;
		}
	}

	public int get(int x) {
		int y = x;
		while (y != p[y]) {
			y = p[y];
		}
		// 压缩
		while (x != p[x]) {
			int t = p[x];
			p[x] = y;
			x = t;
		}
		return y;
	}

	public boolean union(int a, int b) {
		a = get(a);
		b = get(b);
		p[a] = b;
		return a != b;
	}

	public int[] count(){
		int[] a = new int[p.length];
		for(int i = 0;i<a.length;i++){
			a[get(i)]++;
		}
		return a;
	}
	
	public ArrayList<Integer> getRoots(){
		ArrayList<Integer> res = new ArrayList<Integer>();
		for(int i = 0;i<p.length;i++){
			if(get(i)!=i)
				continue;
			res.add(i);
		}
		return res;
	}
}
