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 long long INF = 1 << 28;
long long gcd(long long m, long long n) {
  long long r;
  while (n != 0) {
    r = m % n;
    m = n;
    n = r;
  }
  return m;
}
long long expmod(long long a, long long b) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y) % 1000000007;
    y = (y * y) % 1000000007;
    b >>= 1;
  }
  return (x) % 1000000007;
}
long long inv(long long n) { return expmod(n, 1000000007 - 2); }
int p[200005], b[200005], color[200005];
int main() {
  int n;
  cin >> n;
  memset(color, 0, sizeof(color));
  int ans = 0;
  for (int i = 0; i < n; i++) cin >> p[i];
  int s = 0;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    s += b[i];
  }
  if (s % 2 == 0) {
    ans++;
  }
  int cycles = 0;
  for (int i = 0; i < n; i++) {
    if (color[i] == 0) {
      cycles++;
    }
    int pos = i;
    while (color[pos] == 0) {
      color[pos] = 1;
      pos = p[pos] - 1;
    }
  }
  if (cycles > 1) ans += (cycles);
  cout << ans << endl;
  return 0;
}
