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.Scanner;

public class YetAnotherCountingProblem {

    static long lcm(long a, long b) {
        long t1 = a;
        long t2 = b;
        while (b != 0) {
            long t = a;
            a = b;
            b = t % b;
        }
        return (t1 * t2) / a;
    }

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t-- > 0) {
            long a = scan.nextInt();
            long b = scan.nextInt();
            long q = scan.nextInt();
            long p = Math.max(a, b);
            long lcm = lcm(a, b);

            while (q-- > 0) {
                long l = scan.nextLong() - 1;
                long r = scan.nextLong();
                long res = r - p + 1;
                long e = (r / lcm) - 1;
                res -= (e * p);
                res -= Math.min((r - ((e + 1) * lcm)) + 1, p);

                long res2 = l - p + 1;
                long e2 = (l / lcm) - 1;
                res2 -= (e2 * p);
                res2 -= Math.min((l - ((e2 + 1) * lcm)) + 1, p);
                res -= res2;

                System.out.print(Math.max(0, res) + " ");
            }
            System.out.println();
        }

    }
}