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 MOD = 1000000007;
const int N = 200005;
const double PI = 4 * atan(1);
const long long MAX = 9223372036854775807;
set<char> vowel = {'A', 'O', 'Y', 'E', 'U', 'I', 'a', 'o', 'e', 'u', 'i'};
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool issquare(long long w) { return trunc(sqrt(w)) * trunc(sqrt(w)) == w; }
bool isprime(long long u) {
  for (long long i = 2; i <= (int)sqrt(u); i++) {
    if (u % i == 0) return 0;
  }
  return 1;
}
long long mod(long long to_mod) {
  to_mod %= MOD;
  while (to_mod < 0) to_mod += MOD;
  return to_mod % MOD;
}
long long moduloMultiplication(long long a, long long b, long long mod) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    b >>= 1;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long n, m, x, y, res, z, t, k, ans = 0;
long long a[N];
bool b[N], vis[N];
void dfs(long long src) {
  vis[src] = 1;
  if (vis[a[src]]) return;
  dfs(a[src]);
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(vis, 0, sizeof(vis));
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) cin >> b[i];
  long long cc = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    cc++;
    dfs(i);
  }
  if (cc > 1)
    ans = cc;
  else
    ans = 0;
  res = 0;
  for (int i = 0; i < n; i++) {
    res += b[i];
  }
  ans += 1 - res % 2;
  cout << ans;
  return 0;
}
