You are given two permutations p and q, consisting of n elements, and m queries of the form: l1, r1, l2, r2 (l1 ≤ r1; l2 ≤ r2). The response for the query is the number of such integers from 1 to n, that their position in the first permutation is in segment [l1, r1] (borders included), and position in the second permutation is in segment [l2, r2] (borders included too).

A permutation of n elements is the sequence of n distinct integers, each not less than 1 and not greater than n.

Position of number v (1 ≤ v ≤ n) in permutation g1, g2, ..., gn is such number i, that gi = v.

Input

The first line contains one integer n (1 ≤ n ≤ 106), the number of elements in both permutations. The following line contains n integers, separated with spaces: p1, p2, ..., pn (1 ≤ pi ≤ n). These are elements of the first permutation. The next line contains the second permutation q1, q2, ..., qn in same format.

The following line contains an integer m (1 ≤ m ≤ 2·105), that is the number of queries.

The following m lines contain descriptions of queries one in a line. The description of the i-th query consists of four integers: a, b, c, d (1 ≤ a, b, c, d ≤ n). Query parameters l1, r1, l2, r2 are obtained from the numbers a, b, c, d using the following algorithm: 

  1. Introduce variable x. If it is the first query, then the variable equals 0, else it equals the response for the previous query plus one. 
  2. Introduce function f(z) = ((z - 1 + x) mod n) + 1. 
  3. Suppose l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)), r2 = max(f(c), f(d)). 

Output

Print a response for each query in a separate line.

Examples

Input

3
3 1 2
3 2 1
1
1 2 3 3


Output

1


Input

4
4 3 2 1
2 3 4 1
3
1 2 3 4
1 3 2 1
1 4 2 3


Output

1
1
2
1
1
1
1
1 1 1 1

1

import static java.lang.Math.min;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class C {

	static void solve() throws IOException {
		int testCases = nextInt();
		while (testCases-- > 0) {
			int n = nextInt(), m = nextInt();
			dp.clear();
			long answer = solve(n, m);
			out.println(answer);
		}
	}

	static class Pair {
		int n, m;

		public Pair(int n, int m) {
			this.n = n;
			this.m = m;
		}

		@Override
		public int hashCode() {
			final int prime = 239;
			int result = 1;
			result = prime * result + m;
			result = prime * result + n;
			return result;
		}

		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			Pair other = (Pair) obj;
			if (m != other.m)
				return false;
			if (n != other.n)
				return false;
			return true;
		}
	}

	static Map<Pair, Long> dp = new HashMap<Pair, Long>();

	private static long solve(int n, int m) {
		while ((n & 1) == 0) {
			n >>= 1;
		}
		while ((m & 1) == 0) {
			m >>= 1;
		}
		if (n == 1 && m == 1) {
			return 0;
		}
		Pair pair = new Pair(n, m);
		Long have = dp.get(pair);
		if (have != null) {
			return have;
		}
		long best = Long.MAX_VALUE;
		if (n > 1) {
			best = min(best, solve(n - 1, m) + m);
		}
		if (m > 1) {
			best = min(best, solve(n, m - 1) + n);
		}
		dp.put(pair, best);

		return best;
	}

	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		InputStream input = System.in;
		PrintStream output = System.out;
		File file = new File("c.in");
		if (file.exists() && file.canRead()) {
			input = new FileInputStream(file);
		}
		br = new BufferedReader(new InputStreamReader(input));
		out = new PrintWriter(output);
		solve();
		out.close();
	}

	static long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	static int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	static String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}
}
