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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
const long double eps = 1e-8;
const long double pi = 3.1415926536;
const long double phi = (1 + (long double)sqrt(5)) / 2;
const long double resphi = 2 - phi;
vector<int> color;
vector<int> g;
int num = 0;
int ans;
void dfs(int v) {
  color[v] = 1;
  int to = g[v];
  if (color[to] == 1) {
    num++;
    return;
  }
  if (color[to] == 0) dfs(to);
  color[v] = 2;
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  color.resize(n, 0);
  for (int(i) = 0; (i) < (int)(n); ++(i)) {
    cin >> g[i];
    g[i]--;
  }
  int b = 0;
  for (int(i) = 0; (i) < (int)(n); ++(i)) {
    int temp;
    cin >> temp;
    b += temp;
  }
  ans = 0;
  if (b % 2 == 0) ans++;
  for (int(i) = 0; (i) < (int)(n); ++(i))
    if (color[i] == 0) dfs(i);
  if (num > 1) ans += num;
  cout << ans << endl;
  return 0;
}
