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.util.*;

public class ProbC {
	public static void main(String[] args) {
		ProbC sol = new ProbC();
		sol.start();
	}

	public void start(){
		submit();
	}
	
	public void submit(){
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] p = new int[n];
		int[] b = new int[n];
		for(int i=0;i<n;i++) p[i] = scan.nextInt();
		for(int i=0;i<n;i++) b[i] = scan.nextInt();
		scan.close();
		
		int ret = skewers(p, b);
		System.out.println(ret);
		System.out.flush();
	}
	
	public int skewers(int[] p, int[] b){
		int ret = 1;
		for(int i : b) ret += i;
		ret %= 2;
		
		int n = p.length;
		for(int i=0;i<n;i++) p[i] -= 1;
		
		n = getCycles(p).length;
		if(n==1) n = 0;
		return ret + n;
	}
	
	public int[] getCycles(int[] nums){
		int n = nums.length;
		boolean[] mark = new boolean[n];
		Arrays.fill(mark, false);
		LinkedList<Integer> cycles = new LinkedList();
		for(int i=0;i<n;i++){
			if(mark[i]) continue;
			mark[i] = true;
			int size = 1;
			int j = nums[i];
			while(j!=i){
				mark[j] = true;
				j = nums[j];
				size += 1;
			}
			cycles.add(new Integer(size));
		}
		
		int[] ret = new int[cycles.size()];
		int count = 0;
		for(Integer i : cycles) ret[count++] = i;
		Arrays.sort(ret);
		return ret;
	}
}