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;
const int N = 2e5 + 10;
int n, m, a, b[N], low[N], idx[N], dfs = 1, comp[N], cmp = 0, in[N], mark[N],
                                   p[N];
bool vis[N];
vector<int> S;
vector<vector<int> > g, SCC;
void DFS(int u) {
  idx[u] = low[u] = dfs++;
  vis[u] = true;
  S.push_back(u);
  for (int i = 0; i < g[u].size(); ++i) {
    if (idx[g[u][i]] == 0) DFS(g[u][i]);
    if (vis[g[u][i]]) low[u] = min(low[u], low[g[u][i]]);
  }
  if (low[u] == idx[u]) {
    SCC.push_back(vector<int>());
    int flag = 0;
    while (true) {
      int v = S.back();
      flag |= b[v];
      S.pop_back();
      vis[v] = false;
      SCC.back().push_back(v);
      comp[v] = cmp;
      if (v == u) break;
    }
    mark[cmp] = flag;
    ++cmp;
  }
}
int main() {
  scanf("%d", &n);
  g.resize(n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    --x;
    g[i].push_back(x);
  }
  for (int i = 0; i < n; ++i) scanf("%d", b + i);
  for (int i = 0; i < n; ++i)
    if (idx[i] == 0) DFS(i);
  int ans = cmp > 1 ? cmp : 0;
  int cnt = 0;
  for (int i = 0; i < n; ++i) cnt += b[i];
  ans += !(cnt & 1);
  printf("%d", ans);
  return 0;
}
