#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool maxup(T& a, T&& b) {
  if (a < b) {
    a = b;
    return true;
  };
}
template <typename T>
bool maxup(T& a, T& b) {
  if (a < b) {
    a = b;
    return true;
  };
}
template <typename T>
bool minup(T& a, T&& b) {
  if (a > b) {
    a = b;
    return true;
  };
}
template <typename T>
bool minup(T& a, T& b) {
  if (a > b) {
    a = b;
    return true;
  };
}
using VV = vector<vector<long long>>;
using V = vector<long long>;
using P = pair<long long, long long>;
using IP = pair<long long, P>;
template <typename T>
inline void input(vector<T>& v) {
  for (auto& x : v) cin >> x;
}
V nexts;
V visited;
void dfs(long long x) {
  if (visited[x]) return;
  visited[x] = true;
  dfs(nexts[x]);
}
signed main() {
  long long n;
  cin >> n;
  nexts.resize(n);
  visited.resize(n);
  long long bs = 0;
  for (signed i = signed(0); i < signed(n); i++) {
    cin >> nexts[i];
    nexts[i]--;
  }
  for (signed i = signed(0); i < signed(n); i++) {
    long long x;
    cin >> x;
    bs += x;
  }
  long long num = 0;
  for (signed i = signed(0); i < signed(n); i++)
    if (!visited[i]) {
      num++;
      dfs(i);
    }
  long long ans = bs % 2 == 0;
  if (num != 1) {
    ans += num;
  }
  cout << ans << endl;
}

### 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;
template <typename T>
bool maxup(T& a, T&& b) {
  if (a < b) {
    a = b;
    return true;
  };
}
template <typename T>
bool maxup(T& a, T& b) {
  if (a < b) {
    a = b;
    return true;
  };
}
template <typename T>
bool minup(T& a, T&& b) {
  if (a > b) {
    a = b;
    return true;
  };
}
template <typename T>
bool minup(T& a, T& b) {
  if (a > b) {
    a = b;
    return true;
  };
}
using VV = vector<vector<long long>>;
using V = vector<long long>;
using P = pair<long long, long long>;
using IP = pair<long long, P>;
template <typename T>
inline void input(vector<T>& v) {
  for (auto& x : v) cin >> x;
}
V nexts;
V visited;
void dfs(long long x) {
  if (visited[x]) return;
  visited[x] = true;
  dfs(nexts[x]);
}
signed main() {
  long long n;
  cin >> n;
  nexts.resize(n);
  visited.resize(n);
  long long bs = 0;
  for (signed i = signed(0); i < signed(n); i++) {
    cin >> nexts[i];
    nexts[i]--;
  }
  for (signed i = signed(0); i < signed(n); i++) {
    long long x;
    cin >> x;
    bs += x;
  }
  long long num = 0;
  for (signed i = signed(0); i < signed(n); i++)
    if (!visited[i]) {
      num++;
      dfs(i);
    }
  long long ans = bs % 2 == 0;
  if (num != 1) {
    ans += num;
  }
  cout << ans << endl;
}

```