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.
//Codeforces Round 393 (Div. 2)
import java.io.*;
import java.util.*;

public class TaskC {
	
	public static void main (String[] args) throws IOException {
		FastScanner fs = new FastScanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = fs.nextInt();        
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = fs.nextInt()-1;
        }        
        int c = 0;
        for (int i = 0; i < n; i++) {
            c += fs.nextInt();
        }
        int[] b = new int[n];
		int l = 0;
		for (int i = 0; i < n; i++) {
			if (b[i] == 0) {
				l++;
				b[i] = 1;
				int j = a[i];
				while (b[j] == 0) {
					b[j] = 1;
					j = a[j];
				}
			}
		}		
		if (l == 1) l = 0;
		if ((c & 1) == 0) l++;
        pw.println(l);
        		
		pw.close();		
	}

	static class FastScanner {
        BufferedReader reader;
        StringTokenizer tokenizer;        
        FastScanner(InputStream i) {
            reader = new BufferedReader(new InputStreamReader(i));
            tokenizer = new StringTokenizer("");
        }
        String next() throws IOException {
            while(!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(reader.readLine());
            return tokenizer.nextToken();
        }
        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }        
    }
}