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.lang.*;
import java.util.*;
import java.io.*;

public class Main {
    void solve() {
       int n=ni();
       int p[]=new int[n+1];
       int b[]=new int[n+1];
       int vis[]=new int[n+1];
       for(int i=1;i<=n;i++) p[i]=ni();
       for(int i=1;i<=n;i++) b[i]=ni();
       int ans=1;
       for(int i=1;i<=n;i++) ans^=b[i];
       int cc=0,idx;
       for(int i=1;i<=n;i++){
           if(vis[i]==1) continue;
           vis[i]=1;
           idx=i;
           while(vis[p[idx]]==0){
               idx=p[idx];
               vis[idx]=1;
           }
           cc++;
       }
       if(cc>1)
       ans+=cc;
        pw.println(ans);

    }

    long M = (long)1e9+7;
    // END
    PrintWriter pw;
    StringTokenizer st;
    BufferedReader br;

    void run() throws Exception {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        solve();
        pw.flush();
    }
    public static void main(String[] args) throws Exception {
        new Main().run();
    }

    String ns() {
        while (st == null || !st.hasMoreElements()) {

            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    String nextLine() throws Exception {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            throw new Exception(e.toString());
        }
        return str;
    }

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

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

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