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.*;

public class C {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int T = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		while(T-- > 0) {
			long a = sc.nextInt();
			long b = sc.nextInt();
			long c = a*b;
			long[] count = new long[(int)c];
			long cnt = 0;
			for(int i = 0; i < c; i++) {
				if(i % a % b != i % b % a) {
					cnt++;
				}
				count[i] = cnt;
			}
			int Q = sc.nextInt();
			while(Q-- > 0) {
				long L = sc.nextLong();
				long R = sc.nextLong();
				//calc for L-1
				long ans1 = cnt * ((L-1) / c) + count[(int)((L-1) % c)];
				long ans2 = cnt * (R / c) + count[(int)(R % c)];
				sb.append(ans2-ans1+" ");
			}
			sb.replace(sb.length()-1, sb.length(), "\n");
		}
		PrintWriter pw = new PrintWriter(System.out);
		pw.print(sb.toString());
		pw.flush();
	}
	
	static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }
    }

}