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 maxn = 200000 + 100;
int n;
int p[maxn], b[maxn];
int fa[maxn];
set<int> st;
void Init() {
  for (int i = 1; i <= n; ++i) {
    fa[i] = i;
  }
}
int Find(int x) { return (x == fa[x] ? x : fa[x] = Find(fa[x])); }
void unit(int x, int y) {
  int xx = Find(x);
  int yy = Find(y);
  fa[xx] = yy;
}
int main() {
  ios::sync_with_stdio(false);
  while (scanf("%d", &n) != EOF) {
    st.clear();
    int ans = 1;
    Init();
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &p[i]);
      unit(p[i], i);
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &b[i]);
      ans ^= b[i];
    }
    for (int i = 1; i <= n; ++i) {
      st.insert(Find(i));
    }
    ans += st.size();
    if (st.size() == 1) {
      --ans;
    }
    printf("%d\n", ans);
  }
  return 0;
}
