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

public class C 
{
	 
	public static void main(String[] args)throws IOException 
	{	
		BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb=new StringBuffer();
		
		int t=Integer.parseInt(ob.readLine());
        while(t-->0) 
        {
        	String s[]=ob.readLine().split(" ");
            int a=Integer.parseInt(s[0]);
            int b=Integer.parseInt(s[1]);
            int q=Integer.parseInt(s[2]);
            int LCM=(int)lcm(a,b);
            int cnt=0;
            int sum[]=new int[LCM];
            for (int x=0;x<LCM;x++) 
            {
                if(((x%a)%b)!=((x%b)%a)) 
                {
                    cnt++;
                }
                sum[x]=cnt;
            }
            while(q-->0)
            {	
            	String str[]=ob.readLine().split(" ");
                long l=Long.parseLong(str[0]);
                long r=Long.parseLong(str[1]);
                sb.append(count(r,LCM,sum)-count(l-1,LCM,sum) + " ");
            }
            sb.append('\n');
        }
        System.out.print(sb);
    }
 
    private static long count(long y, int lcm, int[] sum) {
        return y/lcm*sum[lcm-1]+sum[(int)(y%lcm)];
    }
 
    private static long lcm(long a, long b) {
        return a*b/gcd(a,b);
    }
 
    private static long gcd(long a, long b) {
        return b > 0 ? gcd(b,a%b) : a;
    }
	
	
}
