Problem: You are given two integers a and b, and q queries. The i-th query consists of two numbers l_i and r_i, and the answer to it is the number of integers x such that l_i ≤ x ≤ r_i, and ((x mod a) mod b) ≠ ((x mod b) mod a). Calculate the answer for each query.

Recall that y mod z is the remainder of the division of y by z. For example, 5 mod 3 = 2, 7 mod 8 = 7, 9 mod 4 = 1, 9 mod 9 = 0.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow.

The first line of each test case contains three integers a, b and q (1 ≤ a, b ≤ 200; 1 ≤ q ≤ 500).

Then q lines follow, each containing two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^{18}) for the corresponding query.

Output

For each test case, print q integers — the answers to the queries of this test case in the order they appear.

Example

Input


2
4 6 5
1 1
1 3
1 5
1 7
1 9
7 10 2
7 8
100 200


Output


0 0 0 2 4 
0 91 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.StringTokenizer;

public class code1 {
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(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 nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			long a = sc.nextLong();
			long b = sc.nextLong();
			int q = sc.nextInt();
			long prefix[] = new long[(int)(a * b) + 1];
			for (int i = 1; i <= (int)(a * b); i++) {
				if (((i % a) % b) != ((i % b % a))) {
					prefix[i] = (long)1;
				}
			}
			for (int i = 1; i <= (int)(a * b); i++) {
				prefix[i] += prefix[i - 1];
			}
			while (q-- > 0) {
				long l = sc.nextLong();
				long r = sc.nextLong();
				l--;
				long u = a * b;
				long p = (l / u) * prefix[(int)u] + prefix[(int)(l % u)];
				long s = (r / u) * prefix[(int)u] + prefix[(int)(r % u)];
				System.out.print((s - p) + " ");

			}
			System.out.println();




		}
		System.out.println();



	}
}