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.*;

public class c {
	public static void main(String[] args) throws IOException {
		FastScanner stdin = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int t = stdin.nextInt();
		
		for (int z = 0; z < t; z ++)
		{
			long a = stdin.nextInt();
			long b = stdin.nextInt();
			int q = stdin.nextInt();
			
			long[] before = new long[(int)(a*b)];
			for (int i = 1; i < a*b; i ++)
			{
				before[i] = before[i-1];
				if ((i%a)%b==(i%b)%a) before[i] ++;
			}
									
			for (int i = 0; i < q; i ++)
			{
				long l = stdin.nextLong();
				long r = stdin.nextLong();
				long ogl = l;
				
				if (a==b)
				{
					System.out.print(0 + " ");
					continue;
				}
				
				if (r < a*b)
				{
					System.out.print((r-l+1-before[(int)r]+before[(int)l-1]) + " ");
					continue;
				}
				
				long add = 0;
				if (l < a*b)
				{
					add = before[(int)(a*b-1)] - before[(int)l-1]; 
					l = a*b;
				}
				
				long hi = r/(a*b);
				long lo = (l-1)/(a*b);
				
				long over = hi*a*b + Math.max(a, b) - r;
				long under = lo*a*b;
				
				over = Math.max(over, 0);
				
				if (l-under < Math.max(a,b))
					under = Math.max(a, b);
				else
					under = 0;
				
				long ans = (r-ogl+1 - add - (hi-lo)*Math.max(a, b) +over-under);
				
				System.out.print(ans + " ");
			}
			System.out.println();
		}
		
		out.flush();
	}
	
	static class FastScanner {
	    BufferedReader br;
	    StringTokenizer st;
		
	    public FastScanner(InputStream i) {
	        br = new BufferedReader(new InputStreamReader(i));
	        st = new StringTokenizer("");
	    }
				
	    public String next() throws IOException {
	        if(st.hasMoreTokens())
	            return st.nextToken();
	        else
	            st = new StringTokenizer(br.readLine());
	        return next();
	    }

	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	   
	    public long nextLong() throws IOException {
	        return Long.parseLong(next());
	    }
	    
	    public double nextDouble() throws IOException {
	        return Double.parseDouble(next());
	    }
	 }
}
