#include <bits/stdc++.h>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
using namespace std;
vector<int> edge[200005];
int vis[200005];
int flg;
void dfs(int u) {
  vis[u] = 0;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (vis[v] == -1)
      dfs(v);
    else if (vis[v] == 0)
      flg = 1;
  }
  vis[u] = 1;
}
int main() {
  int a, b, c, i, j, k, tc, t;
  int n, m, cnt = 0;
  int ara[200005];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a);
    edge[i].push_back(a);
  }
  for (i = 1; i <= n; i++) scanf("%d", &ara[i]);
  memset(vis, -1, sizeof vis);
  for (i = 1; i <= n; i++) {
    if (vis[i] == -1) {
      flg = 0;
      dfs(i);
      if (flg) cnt++;
    }
  }
  if (cnt == 1) cnt--;
  int cc = 0;
  for (i = 1; i <= n; i++) cc += ara[i];
  cnt += (1 - cc % 2);
  printf("%d\n", cnt);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
using namespace std;
vector<int> edge[200005];
int vis[200005];
int flg;
void dfs(int u) {
  vis[u] = 0;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (vis[v] == -1)
      dfs(v);
    else if (vis[v] == 0)
      flg = 1;
  }
  vis[u] = 1;
}
int main() {
  int a, b, c, i, j, k, tc, t;
  int n, m, cnt = 0;
  int ara[200005];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a);
    edge[i].push_back(a);
  }
  for (i = 1; i <= n; i++) scanf("%d", &ara[i]);
  memset(vis, -1, sizeof vis);
  for (i = 1; i <= n; i++) {
    if (vis[i] == -1) {
      flg = 0;
      dfs(i);
      if (flg) cnt++;
    }
  }
  if (cnt == 1) cnt--;
  int cc = 0;
  for (i = 1; i <= n; i++) cc += ara[i];
  cnt += (1 - cc % 2);
  printf("%d\n", cnt);
  return 0;
}

```