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 atexit, io, sys 
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   
##################################### 
def gcd(a,b):
	return a if b == 0 else gcd(b, a%b)
def lcm(a,b):
	return a * b / gcd(a,b)

def compute(a,b):
	l = lcm(a,b)
	h = [0] * l
	c = 0
	for ii in xrange(l):
		if (ii % a) % b == (ii %b) % a: c +=1
		h[ii] = c
	return h
def f(n,h):
	t = len(h)
	return ((n/t) * h[t-1]) + h[n%t]
nt = int(raw_input())

for _ in xrange(nt):
	a,b,nq = map(int ,raw_input().split())
	h = compute(a,b)
	for __ in xrange(nq):
		l,r = map(int, raw_input().split())
		print r- l +1 - (f(r,h) - f(l-1,h)),
	print
