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 maxn = 2e5 + 5;
const long long mod = 1e9 + 7;
const int alphabet = 26;
const int inf = INT_MAX;
const long long linf = 1e18 + 9;
namespace util {
inline long long cnm(char ch) { return (ch - '0'); }
inline long long cap(char ch) { return (ch - 'a'); }
inline void instr(string& s, long long& n) {
  cin >> s;
  n = s.length();
}
inline void instr(string& s) { cin >> s; }
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
bool min(T a, T b, T c) {
  return std::min(a, std::min(b, c));
}
}  // namespace util
using namespace util;
long long p[maxn];
bool vis[maxn];
long long b;
void dfs(long long node) {
  vis[node] = true;
  if (vis[p[node]] == false) dfs(p[node]);
}
void solve(int tc) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k, m, n;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> p[i];
    p[i] -= 1;
  }
  for (i = 0; i < n; i++) {
    long long x;
    cin >> x;
    if (x) b += 1;
  }
  long long cnt = 0;
  for (i = 0; i < n; i++) {
    if (!vis[i]) {
      dfs(i);
      ++cnt;
    }
  }
  if (cnt == 1) cnt = 0;
  if (b == 0) {
    b++;
    ++cnt;
  }
  if ((b & 1) != 1) cnt += 1;
  cout << cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  int i = 1;
  while (i <= t) {
    solve(i);
    i++;
  }
  return 0;
}
