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;
int p[200005], mark[200005], b[200005];
bool vis[200005];
vector<int> adj[200005];
void dfs(int node) {
  int i;
  if (vis[node] == true) {
    return;
  }
  vis[node] = true;
  for (i = 0; i < adj[node].size(); i++) {
    if (!vis[adj[node][i]]) {
      dfs(adj[node][i]);
    }
  }
}
bool dfs2(int node, int st) {
  int i;
  if (vis[node] == true) {
    if (node == st) {
      return true;
    } else {
      return false;
    }
  }
  vis[node] = true;
  for (i = 0; i < adj[node].size(); i++) {
    return dfs2(adj[node][i], st);
  }
}
int main() {
  int t, n, i, j, k, l, m, cnt1, cnt, ans, flag;
  t = 1;
  while (t--) {
    cin >> n;
    for (i = 1; i <= n; i++) {
      cin >> p[i];
    }
    for (i = 0; i < n; i++) {
      cin >> b[i];
    }
    flag = cnt1 = 0;
    ans = 0;
    for (i = 0; i < n; i++) {
      if (b[i] == 1) {
        ans++;
      }
    }
    if (ans % 2 == 0) {
      cnt1++;
    }
    if (n == 1) {
      cout << cnt1 << endl;
      continue;
    }
    for (i = 1; i <= n; i++) {
      adj[i].push_back(p[i]);
    }
    cnt = 0;
    for (i = 1; i <= n; i++) {
      if (vis[i] == false) {
        dfs(i);
        cnt++;
      }
    }
    if (cnt == 1) {
      for (i = 1; i <= n; i++) {
        vis[i] = false;
      }
      if (!dfs2(1, 1)) {
        cnt1++;
      }
      cout << cnt1 << endl;
      continue;
    }
    cout << cnt + cnt1 << endl;
  }
}
