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:
/*
 * Author: pranay.agra
 * Time: 2020-04-26 10:36:17
*/

import java.io.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Scanner;

public class c {
    public static void main(String[] args) throws IOException {
        Scanner std = new Scanner(System.in);
        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
        StringBuilder strBu = new StringBuilder();        

        int t = std.nextInt();
        while (t-- > 0) {
            int a = std.nextInt();
            int b = std.nextInt();
            int q = std.nextInt();
            long lcm = lcm(a, b);
            long subranges = lcm - Math.max(a, b);
            while (q-- > 0) {
                long ti = std.nextLong();
                long ri = std.nextLong();
                long firstVal = 0;
                long secondVal = 0;
                if (a%b == 0 || b%a == 0) {
                    log.write("0 ");
                } else {
                    ti -= 1;
                    long totalranges = ti/lcm;
                    long rem = (ti % lcm) - (long)(Math.max(a, b)) + 1;
                    if (rem < 0) rem = 0;
                    firstVal += (totalranges*subranges + rem);

                    totalranges = ri/lcm;
                    rem = (ri % lcm) - (long)(Math.max(a, b)) + 1;
                    if (rem < 0) rem = 0;
                    secondVal += (totalranges*subranges + rem);

                    log.write((secondVal-firstVal) + " ");
                }
            }
            log.write("\n");
        }
        log.flush();
	

    }

    private static int gcd(int a, int b) {
        if (a == 0) return b;
        return gcd(b % a, a);
    }

    private static int lcm(int a, int b) {
        return (a*b)/gcd(a, b);
    }
}

/**
Remember to check for overflow (long)
Does number need to be greater than a lower bound?
**/