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;
const double Pi = acos((double)-1);
const long long oo((long long)1e+18 + 100);
template <typename T>
inline ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto &it : v) out << it << ' ';
  return out;
}
template <typename T>
inline istream &operator>>(istream &in, vector<T> &v) {
  for (auto &it : v) in >> it;
  return in;
}
template <typename T1, typename T2>
inline ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
  out << p.first << ' ' << p.second << ' ';
  return out;
}
template <typename T1, typename T2>
inline istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T>
inline ostream &operator<<(ostream &out, const set<T> &p) {
  for (auto &i : p) out << i << ' ';
  return out;
}
template <typename T>
inline istream &operator>(istream &in, T &p) {
  return in >> p;
}
template <typename T>
inline ostream &operator<(ostream &out, const T &p) {
  return out << p;
}
template <typename T1, typename T2>
inline pair<T1 &, T2 &> unzip_pair(T1 &p1, T2 &p2) {
  return pair<T1 &, T2 &>(p1, p2);
}
int snm[212345];
int getV(int v) {
  if (snm[v] != v) snm[v] = getV(snm[v]);
  return snm[v];
}
void merge(int a, int b) { snm[getV(a)] = getV(b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout < fixed < setprecision(7);
  for (int i = 0; i < (212345); i++) snm[i] = i;
  int n;
  cin > n;
  for (int i = 0; i < (n); i++) {
    int a;
    cin > a;
    a--;
    merge(i, a);
  }
  set<int> m;
  for (int i = 0; i < (n); i++) m.insert(getV(i));
  int a = 0;
  for (int i = 0; i < (n); i++) {
    int b;
    cin > b;
    a += b;
  }
  int ans = m.size();
  if (ans == 1) ans = 0;
  if (!(a & 1)) ans++;
  cout < ans;
  return 0;
}
