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.util.*;
public class S{
    public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int len = sc.nextInt();
		for(int i = 0; i < len; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			int mul = a * b;
			int[] arr = new int[mul + 1];
			for(int k = 1; k <= mul; k++) {
				if((k % a % b) != (k % b % a))
					arr[k] = arr[k - 1] + 1;
				else
					arr[k] = arr[k - 1];
			}
			for(int j = 0; j < c; j++) {
				long x = sc.nextLong();
				long y = sc.nextLong();
				int count = 0;
				int left = (int)(x % mul) == 0 ? mul : (int)(x % mul);
				int right = (int)(y % mul) == 0 ? mul : (int)(y % mul);
				count -= arr[left - 1];
				count += arr[right];
				count += (y - right - (x - left) + 1) / mul * arr[mul];
				System.out.println(count);
			}
		}
    }
}