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;
vector<int> g[300000];
int v[300000] = {0};
int x = 0;
int h[300000] = {0};
void dfs(int i) {
  if (v[i] != 1) {
    v[i] = 1;
    if (h[i] >= 2) {
      x += h[i] - 1;
    }
    for (int j = 0; j < g[i].size(); j++) {
      dfs(g[i][j]);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int c = 0;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    g[i].push_back(t);
    g[t].push_back(i);
    h[t]++;
  }
  int t1 = 0, a1 = 0;
  for (int i = 1; i <= n; i++) {
    x = 0;
    if (v[i] != 1) {
      c++;
      t1++;
      dfs(i);
    }
    if (x > 0) {
      c += x - 1;
    }
    a1 = x;
  }
  int s = 0;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    if (t) {
      s++;
    }
  }
  if (s % 2 == 0) {
    c++;
  }
  if (t1 == 1 && a1 == 0) {
    c--;
  }
  cout << c;
}
