#include <bits/stdc++.h>
using namespace std;
int a[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n, m, k;
int q(int x) { return x * (x + 1) / 2; }
int p[200005];
int b[200005];
int u[200005];
int dfs(int v) {
  u[v] = 1;
  if (!u[p[v]])
    return dfs(p[v]) + 1;
  else
    return 1;
}
int solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i + 1]);
  }
  int c = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i + 1]);
    c += b[i + 1];
  }
  if (n == 1) {
    cout << (c + 1) % 2;
    return 0;
  }
  int res = 0;
  int ones = 0;
  for (int i = 1; i <= n; ++i)
    if (!u[i]) {
      int k = dfs(i);
      if (k == 1)
        ++ones;
      else
        ++res;
    }
  if (res == 1 && ones == 0) {
    cout << (c + 1) % 2;
    return 0;
  }
  if (res == 0)
    cout << ones + (c + 1) % 2;
  else if (ones == 0)
    cout << res + (c + 1) % 2;
  else
    cout << res + ones + (c + 1) % 2;
  return 0;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
int a[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n, m, k;
int q(int x) { return x * (x + 1) / 2; }
int p[200005];
int b[200005];
int u[200005];
int dfs(int v) {
  u[v] = 1;
  if (!u[p[v]])
    return dfs(p[v]) + 1;
  else
    return 1;
}
int solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i + 1]);
  }
  int c = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i + 1]);
    c += b[i + 1];
  }
  if (n == 1) {
    cout << (c + 1) % 2;
    return 0;
  }
  int res = 0;
  int ones = 0;
  for (int i = 1; i <= n; ++i)
    if (!u[i]) {
      int k = dfs(i);
      if (k == 1)
        ++ones;
      else
        ++res;
    }
  if (res == 1 && ones == 0) {
    cout << (c + 1) % 2;
    return 0;
  }
  if (res == 0)
    cout << ones + (c + 1) % 2;
  else if (ones == 0)
    cout << res + (c + 1) % 2;
  else
    cout << res + ones + (c + 1) % 2;
  return 0;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```