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.*;

public class C {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(f.readLine());
        while(t-->0){
            StringTokenizer st = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());
            int[] arr = new int[a*b];
            long sum = 0;
            for(int i = 0; i < a*b; i++){
                if((i % a)%b != (i % b) % a){
                    arr[i] = 1;
                    sum++;
                }
            }
            long[] prefs = new long[a*b+1];
            prefs[0] = 0;
            for(int i = 1; i <= a*b; i++){
                prefs[i] = prefs[i-1]+arr[i-1];
            }
            StringBuilder ot = new StringBuilder();
            for(int i = 0; i < q; i++){
                st = new StringTokenizer(f.readLine());
                long l = Long.parseLong(st.nextToken());
                long r = Long.parseLong(st.nextToken());
                long sum1 = 0;
                if(l < (a*b) && r < (a*b)){
                    ot.append(prefs[(int)r+1]-prefs[(int)l]+" ");
                    continue;
                }
                if(l%(a*b) != 0) sum1 = prefs[a*b]-prefs[(int)(l%(a*b))];
                long sum2 = 0;
                if(r%(a*b) != 0){
                    sum2 = prefs[(int)(r%(a*b))+1];
                }
                long lb = (l/(a*b));
                if(l % (a*b) != 0) lb++;
                long rb = (r/(a*b));
                long ans = sum1+sum2+(sum*(rb-lb));
                ot.append(ans+" ");
            }
            out.println(ot.toString().trim());
        }


        out.close();
    }
}