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.Scanner;

public class Test {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner scan = new Scanner(System.in);
		
		int n = scan.nextInt();
		int p[] = new int[n];
		int z[] = new int[n];
		int a[] = new int[n];
		int i,j,temp,count=0,ans=0;
		
		for(i=0;i<n;i++)
		{
			z[i]=0;
			a[i]=0;
		}
		
		for(i=0;i<n;i++)
		{
			p[i]=scan.nextInt();
			p[i]=p[i]-1;
		}
		
		
		for(i=0;i<n;i++)
		{
			temp=scan.nextInt();
			if(temp==1)
			count++;
		}	
		
		if(count%2==0)
		ans++;
		
		z[0]=1;
		i=0;j=0;count=0;
		for(;count!=(n-1);)
		{	
			while(z[j]!=0)
			{
				j++;
				if(j>=n)
				break;
			}
			
			if(z[p[i]]==0)
			{
				z[i]=1;i=p[i];count++;
			}
			
			else if(i!=j)
			{
				temp=p[i];
				p[i]=p[j];
				p[j]=temp;
				a[i]=1;a[j]=1;
			}
			
			else
			{
				z[j]=1;
				while(z[j]!=0)
				{
					j++;
					if(j>=n)
					break;
				}
				temp=p[i];
				p[i]=p[j];
				p[j]=temp;
				a[i]=1;a[j]=1;
			}	
		}
		
		for(i=0;i<n;i++)
			ans=ans+a[i];
		
		System.out.println(ans);
	}

}
