【codeforces 760C】Pavel and barbecue

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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.

【题目链接】:http://codeforces.com/contest/760/problem/C

【题解】



首先.所有的串要经过所有的n个位置;

则需要那个p排列的循环节组成一个环.

如果那个p排列的循环节组成了多个环(>1).需要把那cnt个环合并成一个环.->需要修改cnt个点的p值;

(环的话用个while就能处理出来)

然后是b数组;

贪心点就是;

只有最后翻转的次数(1的个数)为奇数,才能够保证符合要求.

总的翻转次数为奇数.

则从某个点转移位置,每次轮完一遍,回到原位置的时候,就肯定变成另外一面了;

(每个串到一个点之后是一个状态，再轮一遍又到这同一个点的时候,又变成翻转后的状态了(奇数),这就保证了肯定能符合要求->每个串在每个点都能出现两个状态);



【完整代码】

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 2e5+100;

int n,cnt = 0;
int p[MAXN],b[MAXN];
bool flag[MAXN];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    scanf("%d",&n);
    for (int i = 1;i <= n;i++)
        scanf("%d",&p[i]);
    for (int i = 1;i <= n;i++)
        if (!flag[i])
        {
            cnt++;
            flag[i] = true;
            int t = p[i];
            while (!flag[t])
            {
                flag[t] = true;
                t = p[t];
            }
        }
    if (cnt==1)
        cnt = 0;
    int cnt1 =0,cnt0 = 0;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d",&b[i]);
        if (b[i]==0)
            cnt0++;
        else
            cnt1++;
    }
    if (!(cnt1&1))
        cnt++;
    cout << cnt << endl;
    return 0;
}