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 int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
class union_find_tree {
 private:
  static constexpr int uft_N = 200005;
  queue<int> uft_q1;
  vector<int> uft_parent;
  vector<int> uft_num;

 public:
  union_find_tree(int uft_n) {
    uft_parent.assign(uft_n, -1);
    uft_num.assign(uft_n, 0);
  }
  union_find_tree() {
    uft_parent.assign(uft_N, -1);
    uft_num.assign(uft_N, 0);
  }
  int check_parent(int uft_x) {
    if (uft_parent[uft_x] != -1) {
      uft_q1.push(uft_x);
      return check_parent(uft_parent[uft_x]);
    }
    int uft_a;
    while (!uft_q1.empty()) {
      uft_a = uft_q1.front(), uft_q1.pop();
      uft_parent[uft_a] = uft_x;
    }
    return uft_x;
  }
  bool connect(int uft_x, int uft_y) {
    uft_x = check_parent(uft_x), uft_y = check_parent(uft_y);
    if (uft_x == uft_y) return true;
    if (uft_num[uft_x] > uft_num[uft_y]) swap(uft_x, uft_y);
    uft_parent[uft_x] = uft_y;
    uft_num[uft_y] += uft_num[uft_x];
    return false;
  }
};
int main() {
  int n, m;
  int i, j, k;
  int a, b, c;
  union_find_tree ua;
  map<int, int> m1;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a;
    a--;
    ua.connect(i, a);
  }
  b = 0;
  for (i = 0; i < n; i++) {
    a = ua.check_parent(i);
    if (m1.find(a) == m1.end()) b++;
    m1[a] = 1;
  }
  if (b == 1) b--;
  for (i = 0, c = 0; i < n; i++) {
    cin >> a;
    c += a;
  }
  b += 1 - c % 2;
  cout << b << endl;
}
