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 = numeric_limits<long long>::max();
const long long MOD = 1E9 + 7;
const double PI = 2 * acos(0);
template <typename T,
          template <typename ELEM, typename ALLOC = std::allocator<ELEM> >
          class Container>
ostream& operator<<(ostream& o, const Container<T>& container) {
  for (typename Container<T>::const_iterator beg = container.begin();
       beg != container.end(); beg++)
    o << *beg << " ";
  return o;
}
int n;
int MAXN = 1e6;
vector<int> a(MAXN), b(MAXN);
vector<bool> vis(MAXN, false);
vector<vector<int> > g(MAXN);
void dfs(int u) {
  vis[u] = true;
  vector<int> adj = g[u];
  for (int i = 0; i < (int)adj.size(); ++i) {
    if (!vis[adj[i]]) dfs(adj[i]);
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    g[i].push_back(a[i]);
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    if (b[i] == 1) {
      cnt++;
    }
  }
  int cc = 0;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      dfs(i);
      cc++;
    }
  }
  long long ans = cc;
  if (cc == 1) ans--;
  if (cnt % 2 == 0) {
    ans++;
  }
  cout << ans << "\n";
  return 0;
}
