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 final class CF
{
    public static long gcd(long a,long b)
    {
        if(b==0)
        return a;
        return gcd(b,a%b);
    }
    public static long ans(long r,long lcm,long a,long b)
    {
            long y=Math.max(a,b);
            long x=r/lcm;
            x=(x*(lcm-y));
            
            if((r%lcm)>=y)
            x+=(r%lcm)-y+1;
            return x;
    }
     static int K=(int)Math.pow(10,9)+7;
     public static void main(String[]args)throws IOException
	{
        FastReader ob=new FastReader();
        StringBuffer sb=new StringBuffer();
        int T=ob.nextInt();
        while(T-->0)
        {
            long a=ob.nextLong();
            long b=ob.nextLong();
            long q=ob.nextLong();
            long lcm=(a*b)/gcd(a,b);
            while(q-->0)
            {
                long l=ob.nextLong();
                long r=ob.nextLong();
                if(a==b)
                {
                    sb.append("0 ");
                    continue;
                }
                long x=ans(r,lcm,a,b)-ans(l-1,lcm,a,b);
                sb.append(x+" ");
            }
            sb.append("\n");
            
        }
        System.out.println(sb);
	}
}
class Pair
{
	int x,y;
	Pair(int x,int y)
	{
		this.x=x;
		this.y=y;
	}
}
class PairComparator implements Comparator<Pair>
{
	public int compare(Pair a,Pair b)
	{
		return a.x-b.x;
	}
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;
    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e)  {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    public String nextLine()
    {
        String s="";
        try {
        s=br.readLine();
        } catch (IOException e)  {
                e.printStackTrace();
            }
        return s;
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}