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

public class anothercountingprob {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int cases = scan.nextInt();
		for (int i = 0; i < cases; i++) {
			//StringTokenizer st= new StringTokenizer(scan.nextLine());
			int a = scan.nextInt();
			int b = scan.nextInt();
			int[] pattern= new int[a*b];
			int count=0;
			for (int l=0;l<pattern.length;l++) {
				if ((l%a)%b != (l%b)%a) {
					count++;
				}
				pattern[l]=count;
			}
			int q = scan.nextInt();
			for (int j=0;j<q;j++) {
				long l = Long.parseLong(scan.next());
				long r = Long.parseLong(scan.next());
				long cycles=(r-l+1)/((long)a*(long)b);
				System.out.print((pattern[(int)(r%(a*b))] - pattern[((int)(l-1)%(a*b))])+ cycles*count);
				if (j!=q-1)
					System.out.print(' ');
			}
			System.out.println();
		}
		
	}
}
		
