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 + 9;
long long n, m;
long long int j = 1;
int arr[N];
long long a, b, c;
set<int> nodes1[N];
vector<int> nodes2[N];
map<int, int> mp;
queue<int> q;
bool visited1[N];
bool visited2[N];
char w[N];
long long ans1, ans2;
long long arr1[N], arr2[N];
string s, s1;
int k;
int max11 = -1;
long long p = 1;
vector<long long> l;
vector<long long> l1;
int sum = 0;
int path = 0;
int visited[N];
int mod = 1e9 + 7;
int color;
int path1, path2;
bool flag = false;
int skewers = 0;
int mask;
void dfs(int node) {
  mask ^= arr1[node];
  visited[node] = true;
  for (int i = 0; i < nodes2[node].size(); i++) {
    if (visited[nodes2[node][i]] == false) dfs(nodes2[node][i]);
  }
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a);
    nodes2[i].push_back(a);
    nodes2[a].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr1[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (visited[i] == false) {
      skewers++;
      dfs(i);
    }
  }
  if (skewers == 1) skewers = 0;
  cout << skewers + (1 - mask);
}
