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.
#include <bits/stdc++.h>
using namespace std;
bool isPrime(unsigned long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (unsigned long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int pow(long long int a, long long int b, long long int m) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int ncr(int n, int k) {
  if (k > n) return 0;
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
std::vector<int> v[200005];
bool visited[200005];
void dfs(int s) {
  visited[s] = true;
  for (auto i : v[s]) {
    if (!visited[i]) dfs(i);
  }
}
void solve() {
  int n;
  cin >> n;
  int arr[n], b[n];
  for (int i = 0; i < int(n); i++) {
    cin >> arr[i];
  }
  int f = 0;
  for (int i = 0; i < int(n); i++) {
    cin >> b[i];
    f ^= b[i];
  }
  for (int i = 0; i < int(n); i++) {
    v[i + 1].push_back(arr[i]);
    v[arr[i]].push_back(i + 1);
  }
  dfs(1);
  int c = 0, x = 0;
  for (int i = int(1); i <= int(n); i++) {
    if (!visited[i]) {
      dfs(i);
      if (!x) {
        c += 2;
        x = 1;
      } else {
        c++;
      }
    }
  }
  cout << c + (f == 0) << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
}
