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;
vector<int> adj[200005], adj_rev[200005];
vector<bool> vis(200005, false);
vector<int> finished;
int sum[2], cnt;
void dfs(int u) {
  vis[u] = 1;
  for (int v : adj[u]) {
    if (!vis[v]) dfs(v);
  }
  finished.push_back(u);
}
int sccs = 0;
int scc[200005];
void dfs2(int u) {
  vis[u] = 1;
  scc[u] = sccs;
  for (int v : adj_rev[u]) {
    if (!vis[v]) dfs2(v);
  }
  cnt++;
}
int main() {
  int n, m, u, v;
  cin >> n;
  vector<pair<int, int> > edges;
  for (int i = 1; i <= n; i++) {
    cin >> u;
    adj[i].push_back(u);
    adj_rev[u].push_back(i);
    edges.push_back({i, u});
  }
  int summ = 0;
  for (int i = 0; i < n; i++) {
    cin >> u;
    summ += u;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  reverse(finished.begin(), finished.end());
  for (int i = 1; i <= n; i++) vis[i] = false;
  for (int i = 0; i < n; i++) {
    int last = finished[i];
    cnt = 0;
    if (!vis[last]) dfs2(last), sccs++;
    sum[cnt % 2] += cnt;
  }
  if (sccs == 1) {
    if (summ % 2 == 0)
      cout << 1;
    else
      cout << 0;
    return 0;
  }
  if (summ % 2 == 0)
    cout << sccs + 1;
  else
    cout << sccs;
}
