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.Arrays;
	import java.util.Scanner;
	public class roadt
	{
		public static void main (String[] args)
		 {
			Scanner sc=new Scanner(System.in); 
			int t,i;
			t=sc.nextInt();
			while(t>0)
			{
				t--;
				int a=sc.nextInt();
				int b=sc.nextInt();
				int q=sc.nextInt();
				long ans[]=new long[q];
				long l,r;
				int lc=0;
				lc=(int)lcm(a,b);
				int f[]=new int[1000000];
				
					for ( i = 1; i < lc; i ++) {
						f[i] = f[i - 1];
						if(i % a % b != i % b % a)
						                f[i] ++;
					}
					i=0;
					long s=q;
					while(q-- >0) {
						
					l=sc.nextLong();
					r=sc.nextLong();
						ans[i++]= solve(f,r, lc) - solve(f,l - 1, lc);
					}
			for(i=0;i<s;i++)
				System.out.print(ans[i]+" ");
			System.out.println();
				
			}
			
		 }
		  static long gcd(long a, long b) 
		    { 
		    if (a == 0) 
		        return b;  
		    return gcd(b % a, a);  
		    } 
		      
		     
		    static long lcm(long a, long b) 
		    { 
		        return (a*b)/gcd(a, b); 
		    } 
		    
		    static  long solve(int f[],long  p, int n) {
		    	return f[n - 1] * (p / n) + f[(int)(p % n)];
		    }
}