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 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

/**
 * 
 * @author Kajal
 *
 */
public class YetAnotherCountingProblem {

	public static void main(String[] args) {
		int testCases, a, b, q;
		long l, r;
		String input[];
		StringBuilder out = new StringBuilder();
		BufferedReader br = null;
		
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
//			br = new BufferedReader(new FileReader("in.txt"));
			testCases = Integer.parseInt(br.readLine().trim());
			for (int t = 1;t <= testCases;t++)
			{	
				input = br.readLine().split(" ");
				a = Integer.parseInt(input[0]);
				b = Integer.parseInt(input[1]);
				q = Integer.parseInt(input[2]);
				
				Map<Long, Boolean> xAlreadyCalced = new HashMap<>();
				for (int i = 0;i < q;i++) {
					input = br.readLine().split(" ");
					l = Long.parseLong(input[0]);
					r = Long.parseLong(input[1]);
					if (a <= b && l <= a && r < b) {
						out.append(0).append(" ");
						continue;
					} else if (b <= a && l <= b && r < a) {
						out.append(0).append(" ");
						continue;
					} else {
						long count = 0;
						if (a < b && r-b <= l)
							l = b;
						else if (b < a && r-a <= l)
							l = a;
						for (long x = l;x <= r;x++) {
							if (xAlreadyCalced.containsKey(x)) {
								count++;
								continue;
							}
							if ( ((x%a)%b) != ((x%b)%a)) {
								count++;
								xAlreadyCalced.put(x, true);
							}
						}
						out.append(count).append(" ");
					}
				}
				out.append("\n");
			}	
			System.out.println(out);
			
		} catch (NumberFormatException | IOException e) {
			e.printStackTrace();
		} finally {
			try {
				if (br != null)
					br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
	}

}
