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.*;
import java.util.StringTokenizer;
// given a, b and q queries
// each queries consist of 2 integer Li, Ri;
// find the number of integer between Li, Ri such that ((x % a) % b) != ((x % b ) % a);
/**
@author Tran Anh Tai
 Ideas: if (a == b): return 0 directly;
 else WLOG, a > b -> let x = by + d with d is the remainder when x is divided by b;
 hence, the condition is equivalent to (by) not divided by a;
 hence, y is divided by (b / gcd(a, b));
 x = lcm(a, b) * t + d; with d < b;
 //what to do now is calculating the number of x such that by is divided by a; (1)
 Then the answer is the complement of (1) = (total number of number in range [Li, Ri] - (1));
 */
public class AnotherCountingProb {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    static class Task{
        public void solve(InputReader in, PrintWriter out) {
            int test = in.nextInt();
            int a,b,q;
            long r, l;
            for (int i = 0; i < test; i++) {
                a = in.nextInt();
                b = in.nextInt();
                q = in.nextInt();
                int lcm = (a * b) / gcd(a, b);
                for (int j = 0; j < q; j++){
                    l = in.nextLong();
                    r = in.nextLong();
                    long au = 0;
                    for (int re = 0; re < b; re++){
                        au += (cnt(r, re, lcm) - cnt(l - 1, re, lcm));
                    }
                    // the result is the total number - (the number of number in range[l, r] that satisfied the reverse condition);
                    out.print(r - l + 1 - au + " ");
                }
                out.println();
            }
        }
        // count the total number of integer x < l such that (x % a == d);
        public static long cnt(long l, int d, int a){
            if  (l < d){
                return 0;
            }
            int start = 0;
            long end = (l - d) / a;
            return (end - start + 1);
        }
        public static int gcd(int a, int b){
            if (a == 0){
                return b;
            }
            else{
                return gcd(b % a, a);
            }
        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
    }
}
