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;
class UnionFind {
 private:
  vector<int> uf;

 public:
  UnionFind(int n) : uf(n, -1) {}
  int find(int x) {
    if (uf[x] < 0) {
      return x;
    }
    return uf[x] = find(uf[x]);
  }
  bool unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x != y) {
      if (uf[x] > uf[y]) {
        swap(x, y);
      }
      uf[x] += uf[y];
      uf[y] = x;
      return true;
    } else {
      return false;
    }
  }
  int size(int x) { return -uf[find(x)]; }
  bool same(int x, int y) { return find(x) == find(y); }
};
int main() {
  int n;
  cin >> n;
  UnionFind uf(n);
  int cc = n;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    a--;
    cc -= uf.unite(i, a);
  }
  int b = 1;
  for (int i = 0; i < n; i++) {
    int tmp;
    scanf("%d", &tmp);
    b ^= tmp % 2;
  }
  if (cc == 1) cc = 0;
  cout << cc + b << endl;
}
