#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int p[200010], b[200010], v[200010];
int main() {
  int T, i, j, k, ca = 0, m = 0, K = 0, n;
  scanf("%d", &n);
  int s = 0;
  for (int i = 0; i < n; i++) scanf("%d", &p[i]), p[i]--;
  for (int i = 0; i < n; i++) scanf("%d", &b[i]), s += b[i];
  for (int i = 0; i < n; i++)
    if (!v[i]) {
      int first = i;
      while (!v[first]) {
        v[first] = 1;
        first = p[first];
      }
      m++;
    }
  if (m == 1) {
    printf("%d\n", s % 2 == 0 ? 1 : 0);
  } else {
    printf("%d\n", m + (s % 2 == 0 ? 1 : 0));
  }
  return 0;
}

### Prompt

Generate a Cpp solution 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int p[200010], b[200010], v[200010];
int main() {
  int T, i, j, k, ca = 0, m = 0, K = 0, n;
  scanf("%d", &n);
  int s = 0;
  for (int i = 0; i < n; i++) scanf("%d", &p[i]), p[i]--;
  for (int i = 0; i < n; i++) scanf("%d", &b[i]), s += b[i];
  for (int i = 0; i < n; i++)
    if (!v[i]) {
      int first = i;
      while (!v[first]) {
        v[first] = 1;
        first = p[first];
      }
      m++;
    }
  if (m == 1) {
    printf("%d\n", s % 2 == 0 ? 1 : 0);
  } else {
    printf("%d\n", m + (s % 2 == 0 ? 1 : 0));
  }
  return 0;
}

```