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 inf = (int)1e9 + 7;
const int maxn = (long long)4e5 + 7;
const int lmaxn = (int)1e6 + 7;
const long long linf = (long long)1e18 + 7;
long long gcd(long long a, long long b) {
  while (b) a %= b, swap(a, b);
  return a | b;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
bool isprime(long long x) {
  long long q = 0;
  if (x == 1) return false;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
long long n, pos;
long long a[maxn], p[maxn], b[maxn], bb[maxn], u[maxn], sz;
vector<long long> g[maxn];
vector<long long> pv;
void dfs(long long v) {
  u[v] = 1;
  if (u[a[v]] == 0) dfs(a[v]);
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i], p[i] = a[i];
  for (long long i = 1; i <= n; i++) cin >> b[i], bb[i] = b[i], pos += b[i];
  for (long long i = 1; i <= n; i++) {
    if (!u[i]) {
      dfs(i);
      sz++;
    }
  }
  if (sz == 1) sz = 0;
  long long ans = sz + (pos + 1) % 2;
  cout << ans;
  return 0;
}
