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

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t=Integer.parseInt(br.readLine());
        while(t-->0)
        {
           String str[]=br.readLine().split(" ");
           long a=Long.parseLong(str[0]);
           long b=Long.parseLong(str[1]);
           int q=Integer.parseInt(str[2]);
           long p=(a*b)/gcd(a,b);
           while(q-->0)
           {
               long max=Math.max(a,b);
               String mystr[]=br.readLine().split(" ");
               long l=Long.parseLong(mystr[0]);
               long r=Long.parseLong(mystr[1]);
               long start=(l%p==0)?(l/p):(l/p+1);
               long end=r/p;
               long ans=0; 
               
               if(Math.min(max,r+1)-l>0)
               ans=ans+Math.min(max,r+1)-l;
               
               if(end>=start)
               {
                   ans=ans+(end-start)*max;
                   ans=ans+Math.min(r-p*end+1,max);
                  
               }
               
              if(p*(l/p)<l&&(l/p)>0)
               {
                   ans=ans+Math.max(0,Math.min(r+1,p*(l/p)+max)-l);
                   
               }
             
               
               
               
               
               ans=r-l+1-ans;
               bw.write(ans+" ");
               
           }
           bw.newLine();
          
           
        }
        
       
       bw.close();
    }
}