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.ArrayList;
import java.util.Iterator;
import java.util.Scanner;

public class Main {
	
	public ArrayList<Integer> adj[];
	public int n;
	public static int result;
	public int b[];
	public int vis[];
	public static Scanner in;
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Main cd = new Main();
		in = new Scanner(System.in);
		cd.createGraph(in);
		System.out.println(cd.countCycles() + cd.inputSeq(in));
		//System.out.println(result);
	}
	
	public int inputSeq(Scanner in) {
		b = new int[adj.length];
		int cnt = 0;
		for (int i = 0; i < adj.length; i++) {
			b[i]  = in.nextInt();
			if (b[i] == 1) {
				cnt++;
			}
		}
		if (cnt % 2 == 0)
			return 1;
		return 0;
	}

	public void createGraph(Scanner in) {
		n = in.nextInt();
		adj = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			adj[i] = new ArrayList<Integer>();
			adj[i].add(in.nextInt()-1);
		}
	}
	
	public int countCycles() {
		result = 0;
		vis = new int[n];
		for (int i = 0; i < adj.length; i++) {
			vis[i] = 0;
		}
		for (int i = 0; i < adj.length; i++) {
			if (vis[i] == 0) {
				DFS(i);
				result++;
			}
		}
		if(result == 1)
		    result = 0;
		return result;
	}
	
	public void DFS(int i) {
		vis[i] = 1;
		for (Iterator it = adj[i].iterator(); it.hasNext();) {
			Integer v = (Integer)it.next();
			if (vis[v] == 0) {
				DFS(v);
			}
		}	
	}
	

}
