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.Scanner;

/**
 * Spend 5 to 10 minutes analyzing different approaches to solve Don't rush and
 * start coding first solution that comes to mind Right now best strategy is
 * analyze for 20 minutes and code 20 minutes 3 question 2 hour. Remaining 30
 * minutes is miscellaneous Simply coding and coding won't solve problem,
 * analyzing then coding will solve it
 */
public class Codeforces {

	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		while (t-->0) {
			int a = in.nextInt(), b = in.nextInt();
			long p[] = new long[a*b+1];
			build(p, a, b);
			int q = in.nextInt();
			for (int i=0;i<q;i++) {
				long l = in.nextLong(), r = in.nextLong();
				System.out.print(query(l, r, p, a, b) + " ");
			}
			System.out.println();
		}
		in.close();
	}

	private static long query(long l, long r, long[] p, int a, int b) {
		return query(r,p,a,b) - query(l-1, p,a,b);
	}

	private static long query(long l, long[] p, int a, int b) {
		int len = a*b;
		long cnt = l/len;
		int rem = (int)(l%len);
		return p[len]*cnt+p[rem];
	}

	private static void build(long[] p, int a, int b) {
		int len = a*b;
		p[0] = 0;
		for (int i=1;i<=len;i++) {
			p[i] = p[i-1];
			if (i%a%b != i%b%a) p[i]++;
		}
	}
	
}