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.math.*;

public class Main {

	private static FastReader fastReader = new FastReader();
	public static void main(String[] args) {
		Task solver = new Task();
		solver.solve();
	}
 
	static class Task{
		
		private long getGCD(long a, long b){
			if(b==0) return a;
			return getGCD(b, a%b);
		}
		
		private long getCount(long r, long lcm, long b){
			long blocks = r/lcm;
			long count = r;
			long pd = blocks*lcm;
			if(blocks != 0){
				count -= Math.min(b, r-pd + 1);
				count -= (blocks-1)*b;
			}
			return (count-Math.min(r, b-1));
		}
		
		public void solve() {
			int nt = fastReader.nextInt();
			StringBuilder result = new StringBuilder();
			while(nt-- > 0){
				long a = fastReader.nextLong();
				long b = fastReader.nextLong();
				int q = fastReader.nextInt();
				if(a>b){
					long temp = a;
					a = b;
					b = temp;
				}
				StringBuilder ans = new StringBuilder();
				while(q-- > 0){
					long l = fastReader.nextLong();
					long r = fastReader.nextLong();
					long count = 0;
					long lcm = (a*b)/getGCD(a, b);
					count = getCount(r, lcm, b);
					count -= getCount(l-1, lcm, b);
					ans.append(String.valueOf(count)+" ");
				}
				result.append(ans+"\n");
			}
			System.out.println(result);
		}
	}
 
	static class FastReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
 
		public FastReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
			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 String readLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}
	}
}