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 <typename T>
void resize(int n, vector<T> &u) {
  u.resize(n);
}
template <typename Head, typename... Tail>
void resize(int n, Head &H, Tail &...T) {
  resize(n, H);
  resize(n, T...);
}
template <typename T>
void debug_out(T t) {
  cerr << t;
}
template <typename A, typename B>
void debug_out(pair<A, B> &u) {
  cerr << "(" << u.first << " " << u.second << ")";
}
template <typename T>
void debug_out(vector<T> &t) {
  int sz = t.size();
  for (int i = 0; i < sz; i++) {
    debug_out(t[i]);
    if (i != sz - 1) cerr << ", ";
  }
}
template <typename T>
void debug_out(vector<vector<T>> &t) {
  int sz = t.size();
  for (int i = 0; i < sz; i++) {
    debug_out(t[i]);
    if (i != sz - 1) cerr << endl;
  }
}
vector<char> lowchar = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                        'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                        's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
vector<char> hichar = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                       'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                       'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
vector<char> base_10 = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'};
template <typename T>
string to_string(T t) {
  ostringstream ss;
  ss << t;
  return ss.str();
}
long long to_num(string t) {
  long long res = 0;
  for (int i = 0; i < (int)t.length(); i++) {
    res *= 10;
    res += t[i] - '0';
  }
  return res;
}
int n, b, cnt, res;
vector<int> a, was, f;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  resize(n, a, was, f);
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    f[i]--;
  }
  int cur = 0;
  for (int i = 0; i < n; i++) {
    cin >> b;
    cur += b;
  }
  res += 1 - (cur & 1);
  for (int i = 0; i < n; i++) {
    if (was[i]) continue;
    cnt++;
    was[i] = 1;
    int s = i;
    while (!was[f[s]]) {
      s = f[s];
      was[s] = 1;
    }
  }
  res += (cnt > 1 ? cnt : 0);
  cout << res;
  return 0;
}
