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.math.BigInteger;
import java.util.*;

public class C {
    public static void main(String[] args) throws Exception {
        FastScannerC fs = new FastScannerC();
        PrintWriter out = new PrintWriter(System.out);

        int n = fs.nextInt(), one = 0, zeroPos = -1;
        int[] p = new int[n+1];
        for(int i = 1; i <= n; i++) p[i] = fs.nextInt();
        int[] b = new int[n];
        for(int i = 0; i < n; i++) {
            b[i] = fs.nextInt();
            if(b[i] == 0) zeroPos = i;
            if(b[i] == 1) one++;
        }
        int res = 0;
        if(one%2 == 0) {
            res++;
            if(zeroPos < 0)
                b[0] = 0;
            else b[zeroPos] = 1;
        }
        //p must have a cycle that connects all nodes
        int cc = 0;
        boolean[] V = new boolean[n+1];
        for(int i = 1; i <= n; i++) {
            if(!V[i]) {
                cc++;
                V[i] = true;
                int at = p[i];
                while(!V[at]) {
                    V[at] = true;
                    at = p[at];
                }
            }
        }
        if(cc > 1) res+= cc;
        out.println(res);
        out.close();
    }
}
class FastScannerC {
    BufferedReader br;
    StringTokenizer st;

    public FastScannerC() {
        try	{
            br = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer("");
        } catch (Exception e){e.printStackTrace();}
    }

    public String next() {
        if (st.hasMoreTokens())	return st.nextToken();
        try {st = new StringTokenizer(br.readLine());}
        catch (Exception e) {e.printStackTrace();}
        return st.nextToken();
    }

    public int nextInt() {return Integer.parseInt(next());}

    public long nextLong() {return Long.parseLong(next());}

    public double nextDouble() {return Double.parseDouble(next());}

    public String nextLine() {
        String line = "";
        try {line = br.readLine();}
        catch (Exception e) {e.printStackTrace();}
        return line;
    }

    public Integer[] nextIntegerArray(int n) {
        Integer[] a = new Integer[n];
        for(int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for(int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    public char[] nextCharArray() {
        return next().toCharArray();
    }
}

