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 sys
import math
#input = sys.stdin.buffer.readline
def lcm(aa,bb):
	g = max(aa,bb)

	while not ((g%aa==0) and (g%bb==0)):
		g+=1
	return g

for _ in range(int(input())):
	a,b,q = map(int,input().split())
	
	LCM = lcm(a,b)

	greater = max(a,b)
	for _ in range(q):
		l,r = map(int,input().split())


		times = r//LCM - l//LCM + 1

		newr = r//LCM*LCM+greater-1
		newl = l//LCM*LCM

		#print(newl,newr,times)
		
		total = newr - newl + 1
		#print(total)
		total -= times*greater
		#print(total)

		if newr<=r:
			total += r-newr
		#print(total)
		if newl+greater<=l:
			total -= (l - (greater+newl)) 
		print(total,end=" ")
		#print()




	print()


