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 long long int MOD = 1e9 + 7;
const long double PI = 3.14159265;
long long int powerWithMod(long long int base, long long int exponent,
                           long long int modulus = LLONG_MAX) {
  long long int result = 1;
  base %= modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent >>= 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long int modInverse(long long int a, long long int m = MOD) {
  return powerWithMod(a, m - 2, m);
}
void stringToUpper(string& strToConvert) {
  transform(strToConvert.begin(), strToConvert.end(), strToConvert.begin(),
            ::toupper);
}
void stringToLower(string& strToConvert) {
  transform(strToConvert.begin(), strToConvert.end(), strToConvert.begin(),
            ::tolower);
}
int n, parent[212345], comp, x, ans, cnt;
void init() {
  for (int i = 0; i < 200001; i++) parent[i] = i;
}
int find(int x) {
  if (x == parent[x]) return x;
  parent[x] = find(parent[x]);
  return parent[x];
}
void unite(int x, int y) {
  int p1 = find(x);
  int p2 = find(y);
  if (p1 == p2) return;
  parent[p1] = parent[p2];
  comp--;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  init();
  cin >> n;
  comp = n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    unite(x - 1, i);
  }
  ans = (comp > 1) ? comp : 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    cnt += x;
  }
  cout << ans + 1 - (cnt & 1) << "\n";
  return 0;
}
