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.*;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;
import java.math.BigInteger;



public class App {
	public static void main(String[] args) {
		
	    Scanner input=new Scanner(System.in);
		    
		
		long t=input.nextInt();
		
		while(t-->0){
		  
			long a=input.nextLong();
			long b=input.nextLong();
			long c=input.nextLong();
		    
			long p=(a*b);
			long f= gcd(a,b);
			
			long lcmp=p/f;
		    
			long count=0;
			long[] arr = new long[(int)lcmp];
		    
		    for(long i=0; i<lcmp; i++){
		        
		        if(  ((i%a)%b) != ((i%b)%a) ){
		            count++;
		        }
		        arr[(int)i]=count;
		        
		    }
		    
		    for(long i=0; i<c; i++){
		        
		    	long x=input.nextLong();
		    	long y=input.nextLong();
		        
		    	long l=((x-1)/lcmp)*count + arr[(int)((x-1)%lcmp)];
		    	long r=(y/lcmp)*count + arr[(int)(y%lcmp)];
		        
		        System.out.print(r-l+ " ");
		    }
		    System.out.println();
		    
		    
		    
		}
	}
	
	 static long gcd(long a, long b) 
		    { 
		      if (b == 0) 
		        return a; 
		      return gcd(b, a % b);  
		    } 				
		
		
	}