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;
using namespace std::chrono;
const int N = 300009;
const int LOG = 20;
int cycles;
void d(int curr, vector<int> &v, vector<vector<int>> &g) {
  v[curr] = 1;
  for (auto c : g[curr]) {
    if (v[c] == 0) {
      d(c, v, g);
    } else if (v[c] == 1) {
      cycles++;
    }
  }
  v[curr] = -1;
}
void solve() {
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  vector<vector<int>> grev(n);
  vector<int> b(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    g[i].push_back(x - 1);
    grev[x - 1].push_back(i);
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    count += b[i];
  }
  vector<int> v(n, 0);
  vector<int> order;
  cycles = 0;
  for (int i = 0; i < n; i++) {
    if (!v[i]) {
      d(i, v, g);
    }
  }
  if (count == 0) {
    cout << (cycles == 1 ? 1 : cycles + 1);
  } else if (count == 1) {
    cout << (cycles == 1 ? 0 : cycles);
  } else {
    if (count % 2 == 0) {
      cout << (cycles == 1 ? 1 : cycles + 1);
    } else
      cout << (cycles == 1 ? 0 : cycles);
  }
}
int main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<milliseconds>(stop - start);
}
