#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
int arr[N];
bool visited[N];
int cnt;
int arr2[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", arr + i);
    visited[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", arr2 + i);
  }
  cnt = 0;
  int onez = 0;
  int v = 0;
  for (int i = 1; i <= n; ++i) {
    if (!visited[i]) {
      int node = i;
      int lol = 0;
      do {
        ++lol;
        visited[node] = 1;
        node = arr[node];
      } while (node != i);
      if (lol > 1) {
        ++v;
      } else {
        onez += 1;
      }
    }
  }
  if (n == 1) {
    cnt = 0;
  } else if (onez == n) {
    cnt = n;
  } else {
    if (v == 1) {
      cnt = 1 + onez;
    } else {
      cnt = v + onez;
    }
  }
  if (v == 1 && onez == 0) {
    cnt = 0;
  }
  int ones = 0;
  for (int i = 1; i <= n; ++i) {
    ones += arr2[i];
  }
  if (!(ones & 1)) {
    ++cnt;
  }
  cout << cnt << endl;
}

### Prompt

Please create a solution in Cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
int arr[N];
bool visited[N];
int cnt;
int arr2[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", arr + i);
    visited[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", arr2 + i);
  }
  cnt = 0;
  int onez = 0;
  int v = 0;
  for (int i = 1; i <= n; ++i) {
    if (!visited[i]) {
      int node = i;
      int lol = 0;
      do {
        ++lol;
        visited[node] = 1;
        node = arr[node];
      } while (node != i);
      if (lol > 1) {
        ++v;
      } else {
        onez += 1;
      }
    }
  }
  if (n == 1) {
    cnt = 0;
  } else if (onez == n) {
    cnt = n;
  } else {
    if (v == 1) {
      cnt = 1 + onez;
    } else {
      cnt = v + onez;
    }
  }
  if (v == 1 && onez == 0) {
    cnt = 0;
  }
  int ones = 0;
  for (int i = 1; i <= n; ++i) {
    ones += arr2[i];
  }
  if (!(ones & 1)) {
    ++cnt;
  }
  cout << cnt << endl;
}

```