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 YetAnotherCountingProblem {

	
	public static long gcd(long a, long b) 
    { 
        if (a == 0) 
            return b; 
          
        return gcd(b%a, a); 
    } 
	
	public static void main(String[] args) {

		Scanner input = new Scanner(System.in);
		int test= input.nextInt();
		long a,b,que,l,r;

		while(test-->0)
		{
			a=input.nextLong();
			b=input.nextLong();
			que=input.nextLong();
			
			if(a<b)
			{
				long temp=a;
				a=b;
				b=temp;
			}
			long hcf=a*b/(gcd(a,b));
			while(que-->0)
			{
				l=input.nextLong();
				r=input.nextLong();
				long ans=calculate(r,hcf,a)-calculate(l-1,hcf,a);
				System.out.print(ans+" ");
			}
			System.out.println();
		}
			
	}

	private static long calculate(long x, long hcf, long a) {
		long ans=0;
		ans=(x/hcf)*(hcf-a);
		x%=hcf;
		if(x>=a) ans+=(x-a+1);
		return ans;
		
	}
}