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.
import java.io.*;
import java.util.*;


public class cfs393C{

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int N = sc.nextInt();
		UnionFind uf = new UnionFind(N);
		for (int i = 0; i < N; i++)
		{
			int a = sc.nextInt() - 1;
			uf.union(a, i);
		}
		int count = uf.count == 1 ? 0 : uf.count;
		int sum = 0;
		for (int i = 0; i < N; i++)
		{
			sum += sc.nextInt();
			sum %= 2; 
		}
		count += sum == 0 ? 1 : 0;
		System.out.println(count);
	}
	
	public static class UnionFind  {
        int count;
        private int[] id;

        public UnionFind(int numElements)   {
            this.count = numElements;
            id = new int[numElements];
            Arrays.fill(id, -1);
        }

        public int union(int a, int b)  {
            int roota = find(a);
            int rootb = find(b);

            if (roota == rootb) {
                return 0;
            }

            count--;
            if (id[rootb] < id[roota])  {
                id[roota] = rootb;
                return -1;
            }
            else if (id[rootb] > id[roota]) {
                id[rootb] = roota;
            }
            else    {
                id[rootb] = roota;
                id[roota]--;
            }
            return 1;
        }
        public int find(int a)  {
            return id[a] < 0 ? a : (id[a] = find(id[a]));
        }


    }
	
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(Reader in) {
			br = new BufferedReader(in);
		}

		public FastScanner() {
			this(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String readNextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] readIntArray(int n) {
			int[] a = new int[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextInt();
			}
			return a;
		}

		long[] readLongArray(int n) {
			long[] a = new long[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextLong();
			}
			return a;
		}
	}

}
