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 N = 3e6 + 10;
const long long mod = 1e9 + 7;
long long power(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
long long add(long long a, long long b) { return (a % mod + b % mod) % mod; }
long long mul(long long a, long long b) { return (a % mod * b % mod) % mod; }
long long n, k;
const long long K = 210;
vector<long long> a(N);
vector<long long> b(N);
vector<long long> adj[N];
long long vis[N];
long long cnt, ans;
void dfs(long long x) {
  if (vis[x]) return;
  vis[x] = 1;
  for (auto i : adj[x]) dfs(i);
}
int32_t main() {
  long long n;
  cin >> n;
  long long temp = 0;
  a.resize(n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    adj[i].push_back(a[i]);
  }
  b.resize(n + 1);
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
      cnt++;
    }
  }
  if (cnt == 1) cnt = 0;
  for (long long i = 1; i <= n; i++) ans += b[i];
  if (!(ans & 1)) temp++;
  cout << temp + cnt;
}
