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

public class C_YetAnotherCountingProblem {

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int __ = 0; __ < T; __++) {
            final int a = sc.nextInt();
            final int b = sc.nextInt();
            final int q = sc.nextInt();

            final int period = a * b;
            final int[] ab = new int[period];
            int sum = 0;
            for (int i = 0; i < period; i++) {
                if ((i % a) % b != (i % b) % a) {
                    sum++;
                }
                ab[i] = sum;
            }
            // now sum is the total number during one period.

            for (int i = 0; i < q; i++) {
                long l = sc.nextLong();
                long r = sc.nextLong();

                System.out.println(r / period * sum + ab[(int) ((r % period))]
                        - (l - 1) / period * sum - ab[(int) (((l - 1) % period))]
                );
            }
        }
    }
}
