#include <bits/stdc++.h>
using namespace std;
long long n;
vector<long long> adj[200100], cycle;
long long start, counter, ans, visited[200100], color[200100];
void dfs_visit(long long s) {
  visited[s] = 1;
  color[s] = 1;
  for (long long x : adj[s]) {
    if (!visited[x]) {
      counter++;
      dfs_visit(x);
    } else if (color[x] == 1 && start == x) {
      cycle.push_back(counter);
    } else if (color[x] == 1) {
      ans++;
      cycle.push_back(counter);
    }
  }
}
void dfs() {
  ans = 0;
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      start = i;
      counter = 1;
      dfs_visit(i);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(visited, 0, sizeof(visited));
  memset(color, 0, sizeof(color));
  cin >> n;
  long long x;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    x--;
    adj[i].push_back(x);
  }
  long long sum = 0, temp;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    sum += temp;
  }
  dfs();
  if (!cycle.empty()) {
    if (cycle.size() > 1) ans += cycle.size();
  }
  ans += (!(sum % 2));
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long n;
vector<long long> adj[200100], cycle;
long long start, counter, ans, visited[200100], color[200100];
void dfs_visit(long long s) {
  visited[s] = 1;
  color[s] = 1;
  for (long long x : adj[s]) {
    if (!visited[x]) {
      counter++;
      dfs_visit(x);
    } else if (color[x] == 1 && start == x) {
      cycle.push_back(counter);
    } else if (color[x] == 1) {
      ans++;
      cycle.push_back(counter);
    }
  }
}
void dfs() {
  ans = 0;
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      start = i;
      counter = 1;
      dfs_visit(i);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(visited, 0, sizeof(visited));
  memset(color, 0, sizeof(color));
  cin >> n;
  long long x;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    x--;
    adj[i].push_back(x);
  }
  long long sum = 0, temp;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    sum += temp;
  }
  dfs();
  if (!cycle.empty()) {
    if (cycle.size() > 1) ans += cycle.size();
  }
  ans += (!(sum % 2));
  cout << ans << endl;
  return 0;
}

```