就是先看排列p，必须满足其是一个环，才满足题意。就处理出有几个环，然后把它们合起来，答案就是多少。

然后再看序列b，自己稍微画一画就会发现，如果有偶数个1肯定是不行哒，否则，它就会再置换一圈回到它自己的位置的时候，正反面的情况和最初始相同，这样怎么转都没法在每个位置烤两面。

所以两部分的答案加起来就是最后的答案。

A. Pavel and barbecue

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 2nplacements.

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.

 

#include
         
          using namespace std;int p[200100],n,ans;bool b[200100],vis[200100];int main(){	//freopen("a.in","r",stdin);	scanf("%d",&n);	for(int i=1;i<=n;++i)	  scanf("%d",&p[i]);	for(int i=1;i<=n;++i)	  {	  	if(vis[i])	  	  continue;	  	++ans;	  	vis[i]=1;	  	int U=p[i];	  	while(U!=i)	  	  {	  	  	vis[U]=1;	  	  	U=p[U]; 	  	  }	  }	int cnt=0;	for(int i=1;i<=n;++i)	  {	  	scanf("%d",&b[i]);	  	cnt+=b[i];	  }	printf("%d\n",(ans==1 ? 0 : ans)+(cnt%2==0));	return 0;}