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, os
input = sys.stdin.read().split('\n')[::-1].pop
def print(s):
	sys.stdout.write(str(s) + ' ')
def endl():
	sys.stdout.write('\n')

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

t = int(input())
ans = []

for _ in range(t):
	a, b, q = [int(x) for x in input().split()]
	a, b = min(a, b), max(a, b)
	c = a*b//gcd(a, b)

	v = []
	curr = 0
	for i in range(c):
		curr += (i%a%b) != (i%b%a)
		v.append(curr)
	

	for _ in range(q):
		l, r = [int(x) for x in input().split()]
		l -= 1
		val = 0
		val += curr*(r//c) + v[r%c]
		val -= curr*(l//c) + v[l%c]
		print(val)
	endl()
