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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

import java.util.StringTokenizer;

public class Solution {
	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastReader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 

		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	} 

	
	public static void main(String[] args) 
	{ 
		FastReader in=new FastReader(); 
		int t  = in.nextInt();
		StringBuilder st =new StringBuilder();
		while(t-->0)
		{
			long a = in.nextInt();
			long b = in.nextInt();
			long q = in.nextInt();
			long max = Math.max(a, b);
			long min = Math.min(a, b);
			a = max;
			b = min;
			boolean flag = false;
			if(a%b==0)
				flag = true;
			long m = a/b; 
			long div = b/gcd(a,b);
			for(int i = 0;i<q;i++)
			{
				long l = in.nextLong();
				long r = in.nextLong();
				if(flag)
				{
					st.append(0+" ");
					continue;
				}
				long ans = 0;
				long pstart = l/a;
				long mstart = (l-a*pstart)/b;
				long fstart = l-a*pstart-b*mstart;
				if((pstart*a)%b!=0)
				{
					ans += a - mstart*b-fstart;
				}
				
				long pend = r/a;
				long mend = (r-a*pend)/b;
				long fend = r-a*pend-b*mend;
				if(pstart!=pend)
				{
					if((pend*a)%b!=0)
					{
						ans += mend*b + fend+1;
					}
				}
				else {
					ans = 0;
					if((pstart*a)%b!=0)
					{
						ans += r-l+1;
					}
				}
				if(pstart+1<pend) {
				long val = find(pstart+1, pend-1, div);
				ans += val*a;
				}
				st.append(ans+" ");
			}
			st.append("\n");
		}
		System.out.println(st.toString());
	}

	private static long find(long pstart, long pend, long div) {
		long muls = (pstart)/div;
		if(pstart%div==0)muls--;
		long mule = (pend)/div;
		return pend-pstart-(mule-muls)+1;
		
	}

	static long gcd(long a, long b)
	{
		if(b==0)
			return a;
		return gcd(b,a%b);
	}
}