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 static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.StrictMath.max;
import static java.lang.System.exit;
import static java.util.Arrays.sort;
import java.io.*;
import java.util.*;


public class A {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    static int n;
    static ArrayList<Integer> g[];
    static boolean[] vis;
    static void dfs(int x) {
        vis[x] = true;
        for(int edge:g[x])
            if(!vis[edge]) dfs(edge);
    }

    static void solve() throws Exception {
        n = nextInt();
        int bb = 0,ans = 0;
        vis = new boolean[n];
        g = new ArrayList[n];
        for(int i=0;i<n;i++) {
            g[i] = new ArrayList<>();
            vis[i] = false;
        }
        for(int i=0;i<n;i++) {
            int x = nextInt();
            g[i].add(x-1);
        }
        for(int i=0;i<n;i++)
            ans += nextInt();

        ans%=2;
        ans = 1 - ans;
        for(int i=0;i<n;i++) if(!vis[i]) {
            dfs(i);
            bb++;
        }
        if(bb == 1) bb--;
        out.print(ans + bb);

    }

    static int nextInt() throws IOException {
        return parseInt(next());
    }

    static long nextLong() throws IOException {
        return parseLong(next());
    }

    static double nextDouble() throws IOException {
        return parseDouble(next());
    }

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

    public static void main(String[] args) {
        try {
            //in = new BufferedReader(new FileReader("/Users/abdelnacer/Desktop/code/C.in"));
            //out = new PrintWriter(new BufferedWriter(new FileWriter("/Users/abdelnacer/Desktop/code/C.out")));
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            long lStartTime = System.currentTimeMillis();
            solve();
            long lEndTime = System.currentTimeMillis();
            //System.out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}