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.lang.*;
import java.util.*;

public class Main{
    static int a = 0, b = 0, g = 0;
    public static void main(String args[]) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuffer sb = new StringBuffer();
        int t = Integer.parseInt(br.readLine());
        while(t-- > 0){
            String s[] = (br.readLine()).split(" ");
            a = Integer.parseInt(s[0]);
            b = Integer.parseInt(s[1]);
            g = (a * b) / gcd(a, b);
            int q = Integer.parseInt(s[2]);
            for(int i = 0; i < q; i++){
                s = (br.readLine()).split(" ");
                long l = Long.parseLong(s[0]);
                long r = Long.parseLong(s[1]);
                long temp = calc(r);
                long tminus = calc(l - 1);
                sb.append(temp - tminus+" ");
            }
            sb.append("\n");
        }
        System.out.println(sb);
    }
    public static long calc(long x){
        long temp = 0; 
        long div = x / g;
        long rem = x - (div * g);
        if(rem - Math.max(a, b) + 1 > 0) temp += rem - Math.max(a, b) + 1;
        long mul = g - Math.max(a, b);
        temp += div * mul;
        return temp;
    }
    public static int gcd(int x, int y){ 
      if (y == 0) return x; 
      return gcd(y, x % y);  
    }
}