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.io.*;
import java.util.*;
import java.math.*;
import java.util.concurrent.ThreadLocalRandom;
public class Main 
{

 static long N = 100;
 static long CNT = 62;
 static long  INF = 1 << 62;




	
	 static long pr[] = new long[50005];
	 static long si[] = new  long[50005];
    static void merge(long u, long v) 
    {
    	u = get(u);
    	v = get(v);
    	if (si[(int)v] > si[(int)u])
    	{
    		long k = v;
    		v = u;
    		u = k;
    	}
    	pr[(int)v] = u;
    	si[(int)u] += si[(int)v];
    }
	  static long get(long u) {
	      	if(pr[(int)u] == u) return  u;
	      	else return pr[(int)u] = get(pr[(int)u]);
	      }
	 static void init()
	 {
		 
	 int n = 5;
	      	for (int i = 0; i <= n; i++)
	      	{
	      		pr[i] = i;
	      		si[i] = 1;
	      	}
	 }
	
	 static long clc(long c ,long m ,long n , long k)
	 {
		 return k - c * m * (2 * n - 2);
	 }
	 static long call(long n2, long m2 ,long n ,long m ,long k)
	 {
	 	if (n == 1 && true)
	 	{
	 		long r = k / m;
	 		if (m2 <= k - r * m) r = r + 1;
	 		return r;
	 	}
	 	 long r = k / m;
	 	 long c = r / (2 * n - 2); 
	 	 long res =  - 1;
	 	if (n2 != n && n2 != 1)
	 		res =res + 2 * c + 1;
	 else
	 		res = res + c + 1;	
	 	 long l = clc(c , m , n , k);
	 	for (int i = 1; l > 0 && i <= n; i++)
	 	{
	 		for (int j = 1; k > 0 && j <= m; j++,  l = l - 1)
	 		{
	 			if (i == n2 && j == m2)
	 				res = res + 1;
	 		}
	 	}
	 	for (int i =(int) n - 1; l > 0 && i > 0; i++)
	 	{
	 		for (int j = 1; l > 0 && j < m + 1; j++, l = l - 1)
	 		{
 				if (i == n2 && j == m2)
	 				res = res + 1;
	 		}
	 	}
	 	return res;
	 }

	// static String k;


	 static long parsenum(long j, long l)
	 {
		 String k = "";
	 	 long cur = 0;
	 	for (int i = (int) j; i <= l; ++i)
	 	{
	 		cur *= 10;
	 		cur += k.charAt(i) - '0';
	 	}
	 	return cur;
	 }

	 static long gcd(long a, long b) 
		{
			if (b == 0)
			{	
				return a;
			}
			
			else
			{
				 return gcd(b, a % b);
			}
		}
	static boolean pri( int k)
	 { 
		if( k == 1) return false;
	     for(int i = 2; i * i <= k; i++)
	     {

	         if( k % i == 0) return false;
	     }
	     return true;
	 }
	static long m , k;
	 static long ll;
	static long rr;
	static boolean check(long x)
	 {
	 	 long lt = summall(x - 1, ll);
	 	 long rt = summall(x - 1, rr);
	 	if (lt + rt + x <= m)
	 		return true;
	 	return false;
	 }
	static long summall(long  s,  long u)
	 {
	 	if (u == 0)
	 		return 0;
	 	if (u > s)
	 		return summall(s, s);
	 	return (2 * s - u + 1) * u / 2;
	 }
     static long bsearch()
     {
    	 long l = 0, r = m + 1;
			while (l < r - 1)
			{
				 long md = (l + r) / 2;
				if (check(md))
					l = md;
				else
					r = md;
			}
			
    	 return l + 1;
     }
   static  int n;
    static int p[] =new int[200005];
	 static int b[] = new int[200005];
	 static void input()
	 { 
		 
		 Scanner in = new Scanner(System.in);
		 n = in.nextInt();
		  for(int i = 0; i < n; i++)p[i] = in.nextInt();
		  for(int i = 0; i < n; i++)
		  {
			  b[i] = in.nextInt();
			  p[i] = p[i] - 1;
		  }
	 }
	public static void main(String[] args) throws Exception
	{
		int mon[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
		 Scanner in = new Scanner(System.in);
		   PrintWriter out = new PrintWriter(System.out);
		  
		  
		  
		
		  input();
		  boolean used[] = new boolean[n];
		  long ans = 0;
		  long k = 0;
		    for (int i = 0; i < n; i++) 
		    {
		    	
		    	ans =ans + b[i];
		    	k = ans % 2;
		    }
		    ans = 1 - k;

		    long cnt = 0;
		    for (int i = 0; i < n; i++)
		    {
		        if (used[i]) continue;
		  
		        int y = i;
		        used[y] = true;
		        cnt = cnt + 1;
		        y = p[y];
		        while (y != i)
		        {
		        	if( y % 10000 == -1) init();
		            used[y] = true;
		            y = p[y];
		        }
		    }

		    if (cnt > 1) ans =  cnt+ans;
           out.println(ans);
		
		out.close();
		return;
	}
 

}

	
      

	  
	
