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
input = sys.stdin.readline

def One(type):
	return (type(input()))
def List(type):
	return list(map(type,input().split()))
def Str():
	return (list(input()))
def Many(type):
	return map(type,input().split())

def gcd(a,b):
	if b==0:
		return a
	return gcd(b,a%b)

def lcm(a,b):
	return a*b//gcd(a,b)

def check(a,b,l,r):
	ans = 0
	for i in range(l,r+1):
		if i%a%b!=i%b%a:
			ans+=1
	return ans

def solve():
	a,b,q = Many(int)
	num = lcm(a,b)

	if a>b:
		c = a
		a = b
		b = c

	for _ in range(q):
		l,r = Many(int)
		if r<=b:
			print(0)
			continue

		ans = r-max(b,l)+1
		qty = ((r//num)-(max(l-b-1,0)//num))*b

		if l%num<=b and l>=num:
			qty-= b
			qty+=b-l%num
		if r%num<b and r>=num:
			qty-= b
			qty+=r%num+1
			

		print(ans-qty)

def main():
	t = One(int)
	while t>0:
		t-=1
		solve()

main()
