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 class Major {
    static void dfs(int idx,boolean visited[],int array[]){
        visited[idx] = true;
        if(!visited[array[idx]])dfs(array[idx],visited,array);
    }
    public static void main(String[] args)throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        int array[] = new int[n + 1];
        for(int i = 1;i <= n;i++)
            array[i] = Integer.parseInt(st.nextToken());
        boolean visited[]= new boolean[n + 1];
        int var = 0;
        st = new StringTokenizer(br.readLine());
        for(int i = 1;i <= n;i++){
            int temp = Integer.parseInt(st.nextToken());
            if(temp == 1)++var;
        }
        int cnt = 0;
        if(var % 2 == 0)++cnt;
        int cycles = 0;
        for(int i = 1;i <= n;i++){
            if(visited[i])continue;
            ++cycles;
            dfs(i,visited,array);
        }
        if(cycles == 1)cycles = 0;
        System.out.println(cnt + cycles);
    }
}