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.

/**
 * DA-IICT
 * Author : Savaliya Sagar
 */
import java.io.*;
import java.math.*;
import java.util.*;

public class C {

	public static int mod = 1000000007;
	public static FasterScanner in = new FasterScanner();
	public static PrintWriter o = new PrintWriter(System.out);
	public static HashSet<Integer> h[];
	static boolean visited[];
	public static void main(String[] args) {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					Solve();
					o.close();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
		}, "1", 1 << 26).start();
	}

	public static void Solve() {
		int n = in.nextInt();
		h = new HashSet[n+1];
		visited = new boolean[h.length];
		for(int i=0;i<=n;i++){
			h[i] = new HashSet<Integer>();
		} 
		for(int i=1;i<=n;i++){
			int p = in.nextInt();
			h[i].add(p);
		}
		int count = 0;
		for(int i=0;i<n;i++){
			int b = in.nextInt();
			if(b==1){
				count++;
			}
		}
		int com = 0;
		for(int i=1;i<=n;i++){
			if(!visited[i]){
				com++;
				dfs(i);
			}
		}
		int ans = 0;
		if(com!=1){
			ans = com;
		}
		if(count%2==0){
			ans++;
		}
		o.println(ans);
	}
	 public static void dfs(int node){
		        Stack<Integer> s = new Stack<>();
		        s.push(node);
		        visited[node] = true;
		        while (!s.isEmpty()){
		            int current = s.pop();
		            for(int k:h[current]){
		                if(!visited[k])
		                {
		                    s.push(k);
		                    visited[k] = true;
		                }
		            }
		        }

		    }

	public static long gcd(long n1, long n2) {
		long r;
		while (n2 != 0) {
			r = n1 % n2;
			n1 = n2;
			n2 = r;
		}
		return n1;
	}

	public static long pow(long x, long n, long mod) {
		long res = 1;
		x %= mod;
		while (n > 0) {
			if (n % 2 == 1) {
				res = (res * x) % mod;
			}
			x = (x * x) % mod;
			n /= 2;
		}
		return res;
	}

	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	static class FasterScanner {
		private byte[] buf = new byte[1024];
		private int curChar;
		private int snumChars;

		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = System.in.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		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;
		}

		private boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
	}
}
