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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.Scanner;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author David Wu
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            TaskC.FastScanner sc = new TaskC.FastScanner();
            int n = sc.nextInt();
            int[] P = new int[n + 1];
            int[] B = new int[n + 1];
            int s = 1;
            for (int i = 1; i < n + 1; i++) {
                P[i] = sc.nextInt();
            }
            for (int i = 1; i < n + 1; i++) {
                B[i] = sc.nextInt();
                s += B[i];
            }
            s %= 2;
            int loops = 0;
            boolean[] checked = new boolean[n + 1];
            for (int i = 1; i <= n; i++) {
                if (checked[i]) continue;
                loops++;
                int next = i;
                while (!checked[next]) {
                    checked[next] = true;
                    next = P[next];
                }
            }
            if (loops == 1) out.println(s);
            else out.println(loops + s);
        }

        public static class FastScanner {
            BufferedReader br;
            StringTokenizer st;

            public FastScanner() {
                br = new BufferedReader(new InputStreamReader(System.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());
            }

        }

    }
}

