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;
long long maxl(long long aa, long long bb) {
  if (aa < bb)
    return bb;
  else
    return aa;
}
long long minl(long long aa, long long bb) {
  if (aa > bb)
    return bb;
  else
    return aa;
}
double maxd(double aa, double bb) {
  if (aa > bb)
    return aa;
  else
    return bb;
}
double mind(double aa, double bb) {
  if (aa < bb)
    return aa;
  else
    return bb;
}
long long step(long long a, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return step(a * a, n / 2);
  else
    return a * step(a, n - 1);
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
vector<int> A, B, used;
void dfs(int v) {
  used[v] = 1;
  if (used[A[v]] == 0) dfs(A[v]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  A.resize(n + 1);
  B.resize(n + 1);
  used.resize(n + 1);
  int f = 0;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1; i <= n; i++) {
    cin >> B[i];
    if (B[i] == 1) f++;
  }
  int k = 0;
  if (f % 2 == 0) k++;
  int kol = 0;
  for (int i = 1; i <= n; i++) {
    if (used[i] == 0) {
      kol++;
      dfs(i);
    }
  }
  if (kol == 1) kol = 0;
  cout << k + kol;
  return 0;
}
