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

public class C {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int T = Integer.parseInt(in.readLine());
		for (int i = 0; i < T; i++) {
			StringTokenizer token = new StringTokenizer(in.readLine());
			int A = Integer.parseInt(token.nextToken());
			int B = Integer.parseInt(token.nextToken());
			int q = Integer.parseInt(token.nextToken());
			boolean[] mod = new boolean[A * B + 1];
			for (int a = 1; a <= A * B; a++) {
				mod[a] = (a % A) % B != (a % B) % A;
			}
			int[] pre = new int[A * B + 1];
			for (int a = 1; a <= A * B; a++)
				pre[a] = pre[a - 1] + (mod[a] ? 1 : 0);
			for (int a = 0; a < q; a++) {
				token = new StringTokenizer(in.readLine());
				long l = Long.parseLong(token.nextToken());
				long r = Long.parseLong(token.nextToken());
				l--;
				long c1 = pre[(int)(l % (A * B))] + (long) pre[A * B] * (l / (A * B));
				long c2 = pre[(int)(r % (A * B))] + (long) pre[A * B] * (r / (A * B));
				System.out.print(c2 - c1 + (a == q - 1 ? "\n" : " "));
			}
		}
	}
}