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;
template <typename T>
inline T getNext() {
  T s = 1, x = 0, c = getc(stdin);
  for (; c <= 32; c = getc(stdin))
    ;
  if (c == 45) s = -1, c = getc(stdin);
  for (; 48 <= c && c <= 57; x = x * 10 + c - 48, c = getc(stdin))
    ;
  return x * s;
}
const double eps = 1e-9;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)(4e6) + 256;
const long long INFLL = 1e18;
vector<int> g[MAXN];
int cnt, q;
int used[MAXN];
void dfs(int v) {
  used[v] = true;
  for (auto it : g[v])
    if (!used[it]) dfs(it);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int v;
    scanf("%d", &v);
    g[i].push_back(v);
    g[v].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    int b;
    scanf("%d", &b);
    cnt += b;
  }
  for (int i = 1; i <= n; ++i) {
    if (!used[i]) {
      q++;
      dfs(i);
    }
  }
  if (q == 1) --q;
  if (!cnt) {
    ++cnt;
    ++q;
  }
  if (!(cnt & 1)) ++q;
  cout << q << '\n';
  return 0;
}
