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;
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int sv[5000010];
void sieve() {
  for (int i = 2; i <= sqrt(5000010); i++) {
    if (sv[i] == 0) {
      sv[i] = i;
      for (int j = i * i; j < 5000010; j += i) {
        sv[j] = i;
      }
    }
  }
}
double pow_(double a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b % 2 == 1) {
    return pow_(a, b - 1) * a;
  } else {
    double x = pow_(a, b / 2);
    return x * x;
  }
}
long long bin_s(vector<pair<long long, long long> >& v, long long x) {
  long long l = 0, r = v.size() - 1;
  while (l != r - 1) {
    long long m = (l + r) / 2;
    if (v[m].first > x) {
      r = m;
    } else {
      l = m;
    }
  }
  if (v[r].first == x) {
    return r;
  } else if (v[l].first == x) {
    return l;
  } else {
    return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  srand(time(NULL));
  int n, ans = 0, sum = 0;
  cin >> n;
  vector<int> p(n), d(n);
  vector<bool> used(n, false);
  vector<pair<int, int> > cycle;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> d[i];
    sum += d[i];
    sum %= 2;
  }
  for (int i = 0; i < n; i++) {
    if (!used[i]) {
      v.push_back(cycle.size());
      int x = i;
      while (!used[x]) {
        cycle.push_back(pair<int, int>(p[x] - 1, x));
        used[x] = true;
        x = p[x] - 1;
      }
    }
  }
  for (ans = v.size() - 1; ans > 0; ans--) {
    swap(cycle[cycle.size() - 1].first, cycle[v[ans] - 1].first);
  }
  ans = 0;
  sort(cycle.begin(), cycle.end(), comp);
  for (int i = 0; i < n; i++) {
    if (p[i] != cycle[i].first + 1) {
      ans++;
    }
  }
  cout << ans + 1 - sum;
  return 0;
}
