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;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Simone Vuotto
 */
public class C {
    
     public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int sumB = 0;
        int [] p = new int[n];
        boolean [] visited = new boolean[n];
        int [] b = new int[n];
        for(int i = 0; i < n; ++i) {
            p[i] = scanner.nextInt() - 1;
            visited[i] = false;
        }
        for(int i = 0; i < n; ++i) {
            b[i] = scanner.nextInt();
            sumB += b[i];
        }
        
        int nSubGraphs = 0;
        
        for(int i = 0; i < n; i ++) {
            int j = i;
            if(!visited[j]) {
                ++nSubGraphs;
                while(!visited[j]) {
                    visited[j] = true;
                    j = p[j];
                }
            }
        }
        int requiredChanges = nSubGraphs < 2 ? 0 : nSubGraphs;
        
        if(sumB % 2 == 0) {
            ++requiredChanges;
        }
            
        
        System.out.println(requiredChanges);
     }
}
