Pavel and barbecue

Pavel and barbecue

 

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 npositions 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, ..., b_n, consisting of zeros and ones. Each second Pavel move skewer on position i to position p_i, and if b_iequals 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, ..., p_n (1 ≤ p_i ≤ n) — the permutation, according to which Pavel wants to move the skewers.

The third line contains a sequence b₁, b₂, ..., b_n 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.

Example

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.

这题目的意思贼难搞懂

有N个串，在1-N的位置

p[i]表示i位置上的串下一次要去的位置

b[i]=1时，会翻转。

问你最少改变几次才能每个串的正反面在每个位置都出现。

搜索判断环的数量，如果翻转情况为偶数，再加上1。

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
int a[200010];
int vis[200010];
void dfs(int x)
{
    vis[x] = 1;
    if(!vis[a[x]])
        dfs(a[x]);
}
int main()
{
    int n;
    while(~scanf("%d",&n)){
        memset(a,0,sizeof a);
        memset(vis,0,sizeof vis);
        for(int i=1;i<=n;i++){
            scanf("%d",a+i);
        }
        int sum=0;
        for(int i=1;i<=n;i++){
            int x;
            scanf("%d",&x);
            sum += x;
        }
        int ans=0;

        for(int i=1;i<=n;i++){
            if(!vis[i]){
                dfs(i);
                ans++;
                //cout<<i<<endl;
            }
        }
        if(ans == 1)
            ans--;
        if(sum % 2 == 0)
            ans++;
        printf("%d\n",ans);
    }
    return 0;
}