#include <bits/stdc++.h>
using namespace std;
int N;
vector<int> P, B;
int solve() {
  if (N == 1) return B[1] == 0 ? 1 : 0;
  vector<int> seen(N + 1);
  int ncycles = 0;
  for (int u = (1), _b = (N); u <= _b; ++u) {
    if (seen[u]) continue;
    ++ncycles;
    for (int x = u;; x = P[x]) {
      if (seen[x]) break;
      seen[x] = true;
    }
  }
  int res = ncycles == 1 ? 0 : ncycles;
  int n0 = 0, n1 = 0;
  for (int u = (1), _b = (N); u <= _b; ++u) {
    if (B[u] == 0)
      ++n0;
    else
      ++n1;
  }
  if ((n1 % 2) == 0) ++res;
  return res;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  P = vector<int>(N + 1);
  for (int i = (1), _b = (N); i <= _b; ++i) cin >> P[i];
  B = vector<int>(N + 1);
  for (int i = (1), _b = (N); i <= _b; ++i) cin >> B[i];
  int res = solve();
  cout << res << '\n';
  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 N;
vector<int> P, B;
int solve() {
  if (N == 1) return B[1] == 0 ? 1 : 0;
  vector<int> seen(N + 1);
  int ncycles = 0;
  for (int u = (1), _b = (N); u <= _b; ++u) {
    if (seen[u]) continue;
    ++ncycles;
    for (int x = u;; x = P[x]) {
      if (seen[x]) break;
      seen[x] = true;
    }
  }
  int res = ncycles == 1 ? 0 : ncycles;
  int n0 = 0, n1 = 0;
  for (int u = (1), _b = (N); u <= _b; ++u) {
    if (B[u] == 0)
      ++n0;
    else
      ++n1;
  }
  if ((n1 % 2) == 0) ++res;
  return res;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  P = vector<int>(N + 1);
  for (int i = (1), _b = (N); i <= _b; ++i) cin >> P[i];
  B = vector<int>(N + 1);
  for (int i = (1), _b = (N); i <= _b; ++i) cin >> B[i];
  int res = solve();
  cout << res << '\n';
  return 0;
}

```