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;
class TaskC {
 public:
  vector<vector<long long> > g;
  void dfs(long long v) {
    used[v] = 1;
    for (long long i = 0; i < g[v].size(); i++)
      if (!used[g[v][i]]) dfs(g[v][i]);
    return;
  }
  vector<bool> used;
  void solve(std::istream& cin, std::ostream& cout) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long n;
    cin >> n;
    vector<long long> v(n);
    used.resize(n, false);
    g.resize(n);
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
      v[i]--;
      g[i].push_back(v[i]);
      g[v[i]].push_back(i);
    }
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      long long k;
      cin >> k;
      sum += k;
    }
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      if (!used[i]) dfs(i), ans++;
    }
    if (ans == 1) ans = 0;
    cout << ans + (sum + 1) % 2;
  }
};
int main() {
  TaskC solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
