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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.StringTokenizer;

public class code1 {
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new
			                        InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException  e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			long a = sc.nextLong();
			long b = sc.nextLong();
			int q = sc.nextInt();
			long lc = lcm(a, b);
			long count = 0;
			long tmp = (Math.max(a, b) - 1);
			while (q-- > 0) {
				long l = sc.nextLong();
				long r = sc.nextLong();
				if (l < Math.max(a, b) && r >= Math.max(a, b)) {
					count += tmp - l + 1;
				}
				if (count < 0)
					count = 0;
				if (r < lc) {
					System.out.print(count + " ");
					continue;
				}
				long res = r / lc - l / lc;
				long jj = res * 10;
				long tt = r / lc;
				long g = r - lc * tt;
				if (r <= 10) {
					jj = jj - (10 - r);
				}
				long ans = jj;

				ans = jj + count;
				//System.out.println(ans);
				System.out.print(r - l + 1 - ans);

			}
			System.out.println();



		}
	}
	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	// method to return LCM of two numbers
	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}
}