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;
long long arr[2000000] = {0}, vis[300000] = {0};
vector<long long> vec[300000];
string str;
long long mod = 1000000007;
map<long long, long long> mp;
long long fac(long long n) {
  long long i, prod = 1;
  for (i = 1; i <= n; i++) prod = ((i % mod) * (prod % mod)) % mod;
  return prod;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return 0;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
void dfs(long long v) {
  long long i;
  vis[v] = 1;
  for (i = 0; i < vec[v].size(); i++)
    if (vis[vec[v][i]] == 0) dfs(vec[v][i]);
  return;
}
int main() {
  long long x, y, i, j, k, n, m, num, temp, fn,
      cnt1 = 0, cnt0 = 0, d, l, r, flag = 0, sum1 = 0, sum2 = 0, a, b, len1 = 0,
      t;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> num;
    vec[i].push_back(num);
  }
  for (i = 1; i <= n; i++) {
    cin >> temp;
    if (temp == 1) cnt1++;
  }
  if (cnt1 % 2 == 0) sum1++;
  dfs(1);
  for (i = 2; i <= n; i++) {
    if (vis[i] == 0) {
      sum2++;
      dfs(i);
    }
  }
  if (sum2 > 0) sum2++;
  printf("%lld\n", sum2 + sum1);
  return 0;
}
