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 double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
const long long MOD = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  vector<int> p(n, 0), v(n, 0);
  for (int i = 0; i < n; i++) cin >> p[i];
  for (int i = 0; i < n; i++) p[i]--;
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<vector<int>> qs;
  vector<bool> vis(n, true);
  for (int i = 0; i < n; i++) {
    if (vis[i]) {
      vis[i] = false;
      vector<int> q;
      int st = i, curr = p[i];
      q.push_back(i);
      while (curr != st) {
        vis[curr] = false;
        q.push_back(curr);
        curr = p[curr];
      }
      qs.push_back(q);
    }
  }
  int ctr = 0;
  for (int i = 0; i < qs.size(); i++) {
    if (i + 1 < qs.size()) {
      p[qs[i].back()] = qs[i + 1][0];
      ctr++;
    }
  }
  if (qs.size() > 1) {
    ctr++;
    p[qs.back().back()] = qs[0][0];
  }
  int st = 0, curr = p[0], state = 0;
  if (v[0] == 1) state = 1 - state;
  while (curr != st) {
    if (v[curr] == 1) state = 1 - state;
    curr = p[curr];
  }
  if (state == 0) ctr++;
  cout << ctr << "\n";
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  solve();
  return 0;
}
