Pavel and barbecue

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 b ₁, b ₂, …, b _n, consisting of zeros and ones. Each second Pavel move skewer on position i to position p _i, and if b _i 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 p ₁, p ₂, …, p _n (1 ≤ p _i ≤ n) — the permutation, according to which Pavel wants to move the skewers.

The third line contains a sequence b ₁, b ₂, …, b _n 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.

Solution:

#include <bits/stdc++.h>

using namespace std;

const int N = 1234567;

bool was[N];
int p[N];

int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
    scanf("%d", p + i);
    p[i]--;
    was[i] = false;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (was[i]) {
      continue;
    }
    ans++;
    int x = i;
    while (!was[x]) {
      was[x] = true;
      x = p[x];
    }
  }
  if (ans == 1) {
    ans = 0;
  }
  int sum = 0;
  for (int i = 0; i < n; i++) {
    int foo;
    scanf("%d", &foo);
    sum += foo;
  }
  if (sum % 2 == 0) {
    ans++;
  }
  printf("%d\n", ans);
  return 0;
}