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;
template <class T>
bool remin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool remax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
int nxt() {
  int first;
  scanf("%d", &first);
  return first;
}
long long nxtll() {
  long long first;
  scanf("%" PRId64, &first);
  return first;
}
long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
struct pt {
  long long first, second;
  pt() {}
  pt(long long first, long long second) : first(first), second(second) {}
  inline pt operator-(const pt &r) const {
    return pt(first - r.first, second - r.second);
  }
  inline pt operator+(const pt &r) const {
    return pt(first + r.first, second + r.second);
  }
  inline pt operator*(const long double &r) const {
    return pt(first * r, second * r);
  }
  inline long long sqlen() const {
    return abs(first * first + second * second);
  }
  inline bool operator<(const pt &r) const {
    if (first != r.first) return first < r.first;
    return second < r.second;
  }
  inline bool operator==(const pt &r) const {
    return first == r.first && second == r.second;
  }
};
ostream &operator<<(ostream &os, const pt &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
inline long long cross(const pt &l, const pt &r) {
  return l.first * r.second - l.second * r.first;
}
inline long long dot(const pt &l, const pt &r) {
  return l.first * r.first + l.second * r.second;
}
long long pwmod(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
template <typename T>
inline T sqr(T first) {
  return first * first;
}
void solve() {
  int n = nxt();
  vector<int> p(n);
  for (int i = 0; i < (int)(n); i++) {
    p[i] = nxt() - 1;
  }
  vector<int> b(n);
  for (int i = 0; i < (int)(n); i++) {
    b[i] = nxt();
  }
  int cnt = 0;
  vector<char> u(n, 0);
  for (int i = 0; i < (int)(n); i++) {
    if (u[i]) continue;
    int first = i;
    while (!u[first]) {
      u[first] = 1;
      first = p[first];
    }
    ++cnt;
  }
  int s = accumulate((b).begin(), (b).end(), 0);
  if (cnt == 1) cnt = 0;
  if (!(s & 1)) {
    ++cnt;
  }
  cout << cnt << "\n";
}
void prepare() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  prepare();
  int t = 1;
  for (int i = 0; i < (int)(t); i++) {
    solve();
  }
  return 0;
}
