#include <bits/stdc++.h>
using namespace std;
void add_edge(int u, int v, vector<int> adj[]) { adj[u].push_back(v); }
int get_cycles(vector<int> adj[], int V) {
  bool* visited = new bool[V];
  for (int i = 0; i < V; ++i) {
    visited[i] = false;
  }
  int amount_visited = 0;
  int cycles = 0;
  int vertex = 0;
  int next_vertex = 1;
  while (amount_visited < V) {
    while (!visited[vertex]) {
      visited[vertex] = true;
      amount_visited++;
      vertex = adj[vertex][0];
    }
    cycles++;
    while (next_vertex < V && visited[next_vertex]) {
      next_vertex++;
    }
    vertex = next_vertex++;
  }
  return cycles;
}
int main() {
  int n;
  cin >> n;
  vector<int> adj[n];
  for (int i = 0; i < n; ++i) {
    int input;
    cin >> input;
    add_edge(input - 1, i, adj);
  }
  int ones = 0;
  for (int i = 0; i < n; ++i) {
    int input;
    cin >> input;
    if (input == 1) {
      ones++;
    }
  }
  int ans = 0;
  if (ones % 2 == 0) {
    ans++;
  }
  int cycles = get_cycles(adj, n);
  if (cycles > 1) {
    ans += cycles;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void add_edge(int u, int v, vector<int> adj[]) { adj[u].push_back(v); }
int get_cycles(vector<int> adj[], int V) {
  bool* visited = new bool[V];
  for (int i = 0; i < V; ++i) {
    visited[i] = false;
  }
  int amount_visited = 0;
  int cycles = 0;
  int vertex = 0;
  int next_vertex = 1;
  while (amount_visited < V) {
    while (!visited[vertex]) {
      visited[vertex] = true;
      amount_visited++;
      vertex = adj[vertex][0];
    }
    cycles++;
    while (next_vertex < V && visited[next_vertex]) {
      next_vertex++;
    }
    vertex = next_vertex++;
  }
  return cycles;
}
int main() {
  int n;
  cin >> n;
  vector<int> adj[n];
  for (int i = 0; i < n; ++i) {
    int input;
    cin >> input;
    add_edge(input - 1, i, adj);
  }
  int ones = 0;
  for (int i = 0; i < n; ++i) {
    int input;
    cin >> input;
    if (input == 1) {
      ones++;
    }
  }
  int ans = 0;
  if (ones % 2 == 0) {
    ans++;
  }
  int cycles = get_cycles(adj, n);
  if (cycles > 1) {
    ans += cycles;
  }
  cout << ans << endl;
  return 0;
}

```