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;
const double PI = 3.141592653589793238;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
vector<int> vec[200005];
bool vis[200005];
void dfs(int node) {
  vis[node] = true;
  for (int i = 0; i < vec[node].size(); i++) {
    int Node = vec[node][i];
    if (!vis[Node]) {
      dfs(Node);
    }
  }
}
int main() {
  int n, ans, cnt, v;
  ans = cnt = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v);
    vec[i].push_back(v);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v);
    if (v) cnt++;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
      ans++;
    }
  }
  if (ans == 1) ans = 0;
  ans += (cnt & 1 ? 0 : 1);
  printf("%d\n", ans);
  return 0;
}
