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.util.StringTokenizer;

public class EasyCodeforces {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        StringTokenizer stk = new StringTokenizer(in.readLine());
        int t = Integer.parseInt(stk.nextToken());

        while (t-- > 0) {
            stk = new StringTokenizer(in.readLine());
            int a = Integer.parseInt(stk.nextToken());
            int b = Integer.parseInt(stk.nextToken());
            int q = Integer.parseInt(stk.nextToken());

            int product = a * b;
            int[] equals = new int[product];
            for (int i = 0; i < product; i++) {
                int x = (i % a) % b;
                int y = (i % b) % a;
                if (x == y) {
                    equals[i] = 1;
                }
                if (i > 0) {
                    equals[i] += equals[i - 1];
                }
            }

            while (q-- > 0) {
                stk = new StringTokenizer(in.readLine());
                long l = Long.parseLong(stk.nextToken());
                long r = Long.parseLong(stk.nextToken());

                long ans = r - l + 1;
                long equalsBefore = equals[product - 1] * ((l - 1) / product) + equals[(int) ((l - 1) % product)];
                long equalsTill = equals[product - 1] * (r / product) + equals[(int) (r % product)];
                ans -= (equalsTill - equalsBefore);
                out.print(ans + " ");

            }
            out.println();
        }

        out.close();
        in.close();
    }
}