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.
/**
 * Created by Андрей on 22.01.2017.
 */
import java.util.*;
import java.io.*;

public class TaskC {
    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        int n = in.nextInt();
        int[] m = new int[n];
        int[] b = new int[n];
        for (int i = 0; i<n; i++) {
            m[i] = in.nextInt();
        }
        for (int i = 0; i<n; i++) {
            b[i] = in.nextInt();
        }
        boolean[] v = new boolean[n];
        int c = 0;
        int s = 0;
        for (int i = 0; i<n; i++) {
            if (b[i] == 1) {
                s++;
            }
            if (!v[i]) {
                c++;
                int j = i;
                while (true) {
                    j = m[j] - 1;
                    v[j] = true;
                    if (j == i) {
                        break;
                    }
                }
            }
        }
        int res = c;
        if (res == 1) {
            res--;
        }
        if (s % 2 == 0) {
            res+=1;
        }
        System.out.println(res);
    }

    public void run() {
        try {
            //in = new FastScanner(new File(".in"));
            //out = new PrintWriter(new File(".out"));
            in = new FastScanner();

            solve();

            //out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }
        
        FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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());
        }
    }

    public static void main(String[] arg) {
        new TaskC().run();
    }
}
