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 MAXN = 2e6 + 10, inf = 1e18, mod = 1e9 + 7;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2> &p) const {
    auto h1 = std::hash<T1>{}(p.first);
    auto h2 = std::hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
long long pw(long long te, long long tee) {
  if (tee == 0) return 1;
  long long res = pw(te, tee / 2);
  return (tee % 2 ? te * res * res : res * res);
}
long long n, p[MAXN], b[MAXN], ans = 0;
bool flag = true, mark[MAXN];
void Read_input() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> p[i];
  for (long long i = 1; i <= n; i++) {
    cin >> b[i];
    if (b[i]) flag = false;
  }
}
void dfs(long long v) {
  mark[v] = true;
  if (!mark[p[v]]) dfs(p[v]);
}
void Solve() {
  for (long long i = 1; i <= n; i++)
    if (!mark[i]) {
      dfs(i);
      ans++;
    }
  long long cnt = 0;
  for (long long i = 1; i <= n; i++) cnt += b[i];
  flag = (cnt % 2 == 0);
  if (ans == 1) ans = 0;
  ans += flag;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Read_input(), Solve();
  return 0;
}
