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.*;

public class Main {

    static final int N = 100005;
    static int n;
    static int p[] = new int[N * 2];
    static int b[] = new int[N * 2];
    static int seen[] = new int[N * 2];

    static void dfs(int u){
        seen[u] = 1;
        u = p[u];
        if(seen[u] == 0){
            dfs(u);
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        n = in.nextInt();
        for(int i = 1; i <= n; i++){
            p[i] = in.nextInt();
        }

        for(int i = 1; i <= n; i++){
            b[i] = in.nextInt();
            b[i] += b[i - 1];
        }

        int ans = 0;
        if(b[n] % 2 == 0) ans++;

        int cycle = 0;
        for(int i = 1; i <= n; i++){
            if(seen[i] == 0){
                dfs(i);
                cycle++;
            }
        }

        if(cycle > 1){
            ans += cycle;
        }

        System.out.println(ans);

    }

}
