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.*;
import java.util.*;

public class Sol {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);

		int a = sc.nextInt();
		while (a-- > 0) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			int q = sc.nextInt();
			int max = x*y;
			int arr[] = new int[max+1];
			for (int i = 1; i < max+1; i++) {
				if (((i % x) % y) != ((i % y) % x)) {
					arr[i]=arr[i-1]+1;
				}else{
					arr[i]=arr[i-1];
				}
			}
			long sol=0;
			for (int i = 0; i < q; i++) {
				long l = sc.nextLong();
				long k = sc.nextLong();
				long ans = k/max;
				sol=ans*arr[max];
				ans=k%max;
				sol+=arr[(int)ans];
				l--;
				ans = l/max;
				sol-=ans*arr[max];
				ans=l%max;
				sol-=arr[(int)ans];
				System.out.print(sol+" ");
			}
			System.out.println();
		}
	}

	private static long lcm(long l, long m) {
		long gcd = Gcd(l, m);
		return (l * m) / gcd;
	}

	private static long Gcd(long a, long b) {
		if (b == 0)
			return a;
		else
			return Gcd(b, a % b);

	}

	public static int[] readInt(int n, Scanner sc) {
		int arr[] = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		return arr;
	}

	public static long[] readLong(int n, Scanner sc) {
		long arr[] = new long[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		return arr;
	}
}