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.*;

public class Solution {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tokenizer = null;

    public static void main(String[] args) throws IOException {
        new Solution().execute();
    }

    void debug(Object... os) {
        System.out.println(Arrays.deepToString(os));
    }

    int ni() throws IOException {
        return Integer.parseInt(ns());
    }

    long nl() throws IOException {
        return Long.parseLong(ns());
    }

    double nd() throws IOException {
        return Double.parseDouble(ns());
    }

    String ns() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
            tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }

    String nline() throws IOException {
        tokenizer = null;
        return br.readLine();
    }


    //Main Code starts Here
    int totalCases, testnum;


    void execute() throws IOException {
        totalCases = 1;
        for (testnum = 1; testnum <= totalCases; testnum++) {
            if (!input())
                break;
            solve();
        }
    }

    int n, p[], b[], pc[];

    void solve() throws IOException {
        pc = new int[n];
        int i = 0;
        int a = 0;
        for (i = 0; i < n; i++) {
            if (pc[i] == 0) {
                int ip = i;
                do {
                    pc[ip] = i+1;
                    ip = p[ip];
                } while(ip != i);
                a++;
            }
        }
        int bb = 0;
        for (i = 0; i < n; i++) {
            bb += b[i];
        }
        if (a == 1) { a = 0; }
        if (bb % 2 == 0) { a++; }
        System.out.println(a);
    }

    boolean input() throws IOException {
        n = ni();
        p = new int[n];
        for (int i = 0; i < n; i++)
            p[i] = ni() - 1;
        b = new int[n];
        for (int i = 0; i < n; i++)
            b[i] = ni();
        return true;
    }
}