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.PrintWriter;
import java.util.Scanner;

public class YetAnotherCountingProblem {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int T = sc.nextInt();
        for (int i = 0; i < T; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            int q = sc.nextInt();
            int lcm = a * b / gcd(a, b);
            int[] pref = new int[lcm];
            for (int j = 1; j < lcm; j++) {
                if ((j % a) % b != (j % b) % a) {
                    pref[j] = pref[j - 1] + 1;
                } else {
                    pref[j] = pref[j - 1];
                }
            }
            for (int j = 0; j < q; j++) {
                long l = sc.nextLong();
                long r = sc.nextLong();
                int l1 = (int) (l % lcm);
                int r1 = (int) (r % lcm);
                int add = pref[r1];
                int sub = pref[(l1 - 1 + lcm) % lcm];
                r = r - r1;
                if (l1 != 0) {
                    l = l + lcm - l1;
                }
                long mult = ((r - l) / lcm) * pref[lcm - 1];
                mult = mult + add + pref[lcm - 1] - sub;
                out.print(mult + " ");

            }
            out.println();
        }
        out.close();

    }

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

}