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 CF1342C extends PrintWriter {
	CF1342C() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1342C o = new CF1342C(); o.main(); o.flush();
	}

	int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int q = sc.nextInt();
			int m = a / gcd(a, b) * b;
			int[] kk = new int[m + 1];
			for (int r = 0; r < m; r++)
				if (r % a % b != r % b % a)
					kk[r + 1]++;
			for (int cnt = 1; cnt <= m; cnt++)
				kk[cnt] += kk[cnt - 1];
			while (q-- > 0) {
				long l = sc.nextLong();
				long r = sc.nextLong() + 1;
				int l_ = (int) (l % m);
				int r_ = (int) (r % m);
				long kl = l / m * kk[m] + kk[l_];
				long kr = r / m * kk[m] + kk[r_];
				print(kr - kl + " ");
			}
			println();
		}
	}
}