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.
/** This is a solution to the problem Pavel and Barbecue on codeforces
 * 
 * Given a length <em>n</em>, define the <em>state</em> of an object as (i, o), where i = 1..n and o = 0..1.<br>
 * Given a permutation p and a bit vector b of same length n, a <em>move</em> transforms a state (i, o) to (pi(i), o xor b(i)).<br>
 * Compute the total number of changes which have to be made to a given permutation and bit vector,
 * such that an object iterates through all possible states.<br>
 * <br>
 * For details, see:<br>
 * http://codeforces.com/problemset/problem/756/A<br>
 */

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class PavelAndBarbecue {
	
	private int numObjects;
	int[] permutation;
	int[] bitVector;
	
	public static void main(String[] args) throws IOException {
		new PavelAndBarbecue().solve(InputType.SYSTEMIN.getBufferedReader("input.txt"));
	}
	
	private void solve(BufferedReader br) throws IOException {
		readFromStream(br);
		br.close();
		
		/* Idea: 
		 * Determine number k of components of p. k changes have to be made if k > 1, else 0.
		 * One change has to be made to b, if there are is an even amount of 1s in b.  */
		
		int numCycles = computeNumCycles();
		int numChanges = (numCycles > 1) ? numCycles : 0;
		
		int popCount = Arrays.stream(bitVector).sum();
		numChanges += isEven(popCount) ? 1 : 0;
		
		System.out.println(numChanges);	
	}

	private boolean isEven(int number) {
		return number % 2 == 0;
	}

	private int computeNumCycles() {
		Visited visited = new Visited(numObjects); 
		int numCycles = 0;
		for(int i = 0; i < numObjects; ++i) {
			if(visited.not(i)) {
				numCycles++;
				findCycle(i, visited);
			}
		}
		return numCycles;
	}

	private void findCycle(int i, Visited visited) {
		while(visited.not(i)) {
			visited.mark(i);
			i = permutation[i];
		}
	}
	
	private void readFromStream(BufferedReader br) throws NumberFormatException, IOException {
		numObjects = Integer.parseInt(br.readLine());
		permutation = intArrayFromLine(br.readLine());
		for(int i = 0; i < numObjects; ++i) {
			permutation[i]--;
		}
		bitVector = intArrayFromLine(br.readLine());
	}
	
	private int[] intArrayFromLine(String line) {
		return Arrays.stream(line.split(" ")).mapToInt(Integer::parseInt).toArray();
	}
	
	private static class Visited {
		private boolean[] visited;
		
		Visited(int numObjects) {
			visited = new boolean[numObjects];
			Arrays.fill(visited, false);
		}
		
		private void mark(int i) {
			visited[i] = true;
		}
		
		private boolean not(int i) {
			return !visited[i];
		}
	}
	
	private enum InputType {
		FILE {
			@Override
			public BufferedReader getBufferedReader(String filename) throws FileNotFoundException {
				return new BufferedReader(new FileReader("input.txt"));
			}
		}, 
		SYSTEMIN {
			@Override
			public BufferedReader getBufferedReader(String filename) {
				return new BufferedReader(new InputStreamReader(System.in));
			}
		};
		
		public BufferedReader getBufferedReader(String filename) throws FileNotFoundException {
			return null;
		}
	}
}
