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 int N = 200010;
int n, br = 0, z = 0, ans;
int b[N], o[N];
vector<int> V[N];
void dfs(int i) {
  o[i] = 1;
  if (o[V[i][0]]) {
    z = 1;
    return;
  } else
    dfs(V[i][0]);
  if (z == 1) return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    V[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= n; i++) o[i] = 0;
  br = 0;
  for (int i = 1; i <= n; i++)
    if (!o[i]) {
      z = 0;
      br++;
      dfs(i);
    }
  if (br == 1)
    ans = 0;
  else
    ans = br;
  br = 0;
  for (int i = 1; i <= n; i++)
    if (b[i] == 1) br++;
  if (!(br % 2)) ans++;
  printf("%d", ans);
  return 0;
}
