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;
int N, A[202020], B[202020];
template <int um>
class UF {
 public:
  vector<int> par, rank;
  UF() {
    rank = vector<int>(um, 0);
    for (int i = 0; i < um; i++) par.push_back(i);
  }
  int operator[](int x) {
    return (par[x] == x) ? (x) : (par[x] = operator[](par[x]));
  }
  int operator()(int x, int y) {
    if ((x = operator[](x)) == (y = operator[](y))) return x;
    if (rank[x] > rank[y]) return par[x] = y;
    rank[x] += rank[x] == rank[y];
    return par[y] = x;
  }
};
UF<500000> uf;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  for (i = 0; i < (N); i++) cin >> A[i], A[i]--, uf(i, A[i]);
  x = 1;
  for (i = 0; i < (N); i++) cin >> B[i], x ^= B[i];
  int cnt = 0;
  for (i = 0; i < (N); i++)
    if (uf[i] == i) cnt++;
  if (cnt != 1) x += cnt;
  cout << x << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}
