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;
void itval(istream_iterator<string> it) {}
template <typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  itval(++it, args...);
}
const long long int MOD = 1e9 + 7;
template <typename T>
inline void print(T x) {
  cout << x << "\n";
}
template <typename T>
inline void printvec(T x) {
  for (auto a : x) cout << a << ' ';
  cout << '\n';
}
struct custom {
  bool operator()(const pair<pair<int, int>, int> &p1,
                  const pair<pair<int, int>, int> &p2) const {
    if (p1.first.first == p2.first.first)
      return p1.first.second > p2.first.second;
    return p1.first.first < p2.first.first;
  }
};
long long int get_pow(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
const long long int N = 2e5 + 10, inf = 4e18;
std::vector<vector<int> > adj(N);
vector<int> color(N, 0);
int cycles = 0, f = 0;
void dfs(int s) {
  color[s] = 1;
  for (auto x : adj[s]) {
    if (color[x] == 1 && x != s) {
      f = 1;
    } else if (!color[x])
      dfs(x);
  }
  color[s] = 2;
}
void solve() {
  int n;
  cin >> n;
  int x;
  for (int i = 0; i < n; ++i) {
    cin >> x;
    x--;
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  int su = 0;
  for (long long int i = (long long int)0; i < (long long int)(n); i++) {
    cin >> x;
    su += x;
  }
  for (long long int i = (long long int)0; i < (long long int)(n); i++) {
    if (!color[i]) {
      dfs(i);
      cycles++;
    }
  }
  int ans = 0;
  if (cycles == 1) {
    ans = 0;
  } else {
    ans = cycles;
  }
  ans += int(su % 2 == 0);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  clock_t z = clock();
  for (long long int tes = (long long int)0; tes < (long long int)(test);
       tes++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}
