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.*;
import java.util.*;
import java.util.function.IntBinaryOperator;

public class Solver {
    private BufferedReader in;
    private PrintWriter out;

    Solver () {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter(new FileWriter("output.txt"));
        } catch (Exception ex) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
        try {
            solve();
        } catch (Exception ex) {
            ex.printStackTrace();
        } finally {
            try {
                in.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
            out.close();
        }
    }

    public static void main(String[] args) {
        new Solver();
    }

    StringTokenizer st;
    String next() throws IOException {
        while (st == null || !st.hasMoreElements())
            st = new StringTokenizer(in.readLine());
        return st.nextToken();
    }

    String nextLine() throws IOException {
        return in.readLine();
    }

    int nextInt() throws IOException {
        return Integer.valueOf(next());
    }

    long nextLong() throws IOException {
        return Long.valueOf(next());
    }

    void solve() throws IOException {
        int n = nextInt();
        int[] p = new int[n];
        for (int i = 0; i < n; i++) {
            p[i] = nextInt()-1;
        }
        int sm = 0;
        for (int i = 0; i < n; i++) {
            sm += nextInt();
        }
        int result = (1 - sm % 2);
        boolean[] vis = new boolean[n];
        int cycle = 0;
        for (int i = 0; i < n; i++) {
            if (vis[i]) continue;
            int t = i;
            while (!vis[t]) {
                vis[t] = true;
                t = p[t];
            }
            cycle++;
        }
        out.print((cycle > 1? cycle:0)+result);
    }

}
