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.awt.*;
import java.awt.event.*;
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;

public class A {
	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		//int qq = 1;
		int qq = Integer.MAX_VALUE;
		//int qq = readInt();
		for(int casenum = 1; casenum <= qq; casenum++) {
			int n = readInt();
			int[] to = new int[n];
			int numCycles = 0;
			for(int i = 0; i < n; i++) {
				to[i] = readInt()-1;
			}
			int odd = 0;
			for(int i = 0; i < n; i++) {
				odd += readInt();
			}
			if(n == 1) {
				if(odd == 0) {
					pw.println(1);
				}
				else {
					pw.println(0);
				}
				continue;
			}
			boolean[] seen = new boolean[n];
			for(int i = 0; i < n; i++) {
				if(seen[i]) {
					continue;
				}
				numCycles++;
				seen[i] = true;
				int curr = to[i];
				while(!seen[curr]) {
					seen[curr] = true;
					curr = to[curr];
				}
			}
			odd %= 2;
			odd = 1 - odd;
			int ret = 0;
			if(numCycles > 1) {
				ret = numCycles;
			}
			if(odd == 1) {
				ret++;
			}
			pw.println(ret);
		}
		pw.close();
	}

	public static void exitImmediately() {
		pw.close();
		System.exit(0);
	}

	public static int readInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public static double readDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	public static long readLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	public static String nextLine() throws IOException {
		st = null;
		if(!br.ready()) {
			exitImmediately();
		}
		return br.readLine();
	}

	public static String nextToken() throws IOException {
		while(st == null || !st.hasMoreTokens()) {
			if(!br.ready()) {
				exitImmediately();
			}
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
}
