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.*;
public final class round_393_c
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static int[] parent;
	static TreeSet<Integer> ind=new TreeSet<Integer>(),outd=new TreeSet<Integer>();
	
	static int getParent(int u)
	{
		if(u==parent[u])	
		{
			return u;
		}
		else
		{
			int val=getParent(parent[u]);parent[u]=val;return val;
		}
	}
	
	static void merge(int u,int v)
	{
		parent[v]=u;
	}
	
	@SuppressWarnings("unchecked")
    public static void main(String args[]) throws Exception
    {
		int n=sc.nextInt(),val=0;int[] a=new int[n+1],b=new int[n+1];parent=new int[n+1];
		for(int i=1;i<=n;i++)
		{
			a[i]=sc.nextInt();parent[i]=i;
			ind.add(i);outd.add(i);
		}
		for(int i=1;i<=n;i++)
		{
			b[i]=sc.nextInt();val+=b[i];
		}
		int res=0;
		for(int i=1;i<=n;i++)
		{
			int x=getParent(i),y=getParent(a[i]);
			if(x!=y)
			{
				merge(x,y);
				ind.remove(a[i]);outd.remove(i);
			}
			else if(ind.size()==1)
			{
				continue;
			}
			else
			{
				res++;
			}
		}
		val%=2;res+=(1-val);out.println(res);out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}