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.*;
import java.io.*;
import java.lang.*;
import java.math.*;

public class C {
        public static void main(String[] args) throws Exception {
                BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
                PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
                int t = Integer.parseInt(bf.readLine());
                StringTokenizer st;
                long a, b, q, li, ri;
                for (int i = 0; i < t; i++) {
                        st = new StringTokenizer(bf.readLine());
                        a = Long.parseLong(st.nextToken());
                        b = Long.parseLong(st.nextToken());
                        if (a > b) {
                                long tmp = a;
                                a = b;
                                b = tmp;
                        } // a is smaller
                        q = Long.parseLong(st.nextToken());
                        long l = lcm(a, b);
                        StringBuilder sb = new StringBuilder();
                        for (int j = 0; j < q; j++) {
                                st = new StringTokenizer(bf.readLine());
                                li = Long.parseLong(st.nextToken());
                                ri = Long.parseLong(st.nextToken());
                                long cutL = (li / l);
                                long cutR = (ri / l);
                                long equalTo = (cutR - cutL) * b;
                                equalTo -= min(li - cutL * l, b);
                                equalTo += min(ri - cutR * l + 1, b);
                                sb.append(ri - li + 1 - equalTo);
                                if (j != q - 1) {
                                        sb.append(' ');
                                }
                        }
                        out.println(sb.toString());
                }
                out.close();
                System.exit(0);
        }

        public static long lcm(long a, long b) {
                return (a / gcd(a, b)) * b;
        }

        public static long gcd(long a, long b) {
                long t;
                while (a > 0) {
                        t = b % a;
                        b = a;
                        a = t;
                }
                return b;
        }

        public static long min(long a, long b) {
                return a < b ? a : b;
        }

        public static long max(long a, long b) {
                return a > b ? a : b;
        }

        public static long abs(long a) {
                return a > 0 ? a : -a;
        }
}