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.*;
import java.lang.*;
public class Codechef{
    public static void main(String[] args) throws Exception{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int T=Integer.parseInt(br.readLine());
        StringBuffer sb=new StringBuffer();
        while(T-->0){
            String[] str=br.readLine().split(" ");
            int a=Integer.parseInt(str[0]);
            int b=Integer.parseInt(str[1]);
            int q=Integer.parseInt(str[2]);
            int[] arr=new int[a*b];
            for(int i=0;i<a*b;i++)
                if((i%a)%b!=(i%b)%a) arr[i]=1;
            for(int i=1;i<a*b;i++) arr[i]=arr[i-1]+arr[i];
            int t=arr[a*b-1];
            while(q-->0){
                String[] s=br.readLine().split(" ");
                long l=Long.parseLong(s[0])-1;
                long r=Long.parseLong(s[1]);
                long count1=t*(l/(a*b));
                int pos1=(int)(l%(a*b));
                count1+=arr[pos1];
                long count2=t*(r/(a*b));
                int pos2=(int)(r%(a*b));
                count2+=arr[pos2];
                long ans=count2-count1;
                sb.append(ans+" ");
            }
            sb.append("\n");
        }
        System.out.print(sb);
    }
}