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.StringTokenizer;

/**
 * Created by Anton Berezin on 2/18/2017.
 */
public class Barbeque {

	public static void main(String[] args) throws Exception {
		try (FastScanner fastScanner = new FastScanner();
		     PrintWriter writer = new PrintWriter(new BufferedOutputStream(System.out))) {
			int n = fastScanner.nextInt();
			int[] p = new int[n + 1];
			int[] b = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				p[i] = fastScanner.nextInt();
			}
			int ones = 0;
			for (int i = 1; i <= n; i++) {
				b[i] = fastScanner.nextInt();
				ones += b[i];
			}

			boolean[] visited = new boolean[n + 1];
			int visitedCnt = 0;
			int index = 1;
			int result = ones % 2 == 0 ? 1 : 0;
			int cycles = 0;
			while (visitedCnt < n) {
				int curr = index;
				while (!visited[index]) {
					visited[index] = true;
					visitedCnt++;
					index = p[index];
				}
				if (visitedCnt < n) {
					index = curr;
					while (index <= n && visited[index]) {
						index++;
					}
					cycles++;
				}
			}
			result += cycles == 0 ? 0 : cycles + 1;
			System.out.println(result);
		}
	}

	public static class FastScanner implements AutoCloseable {
		BufferedReader bufferedReader;
		StringTokenizer tokenizer;

		public FastScanner() {
			bufferedReader = new BufferedReader(new InputStreamReader(System.in));
		}

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

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

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

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

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

		@Override
		public void close() throws IOException {
			bufferedReader.close();
		}
	}
}
