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 
def gcd(a,b):
	a, b = min(a,b), max(a,b)
	if a == 1:
		return 1
	if b%a == 0:
		return a
	return gcd(a, b%a)

t = int(input())
for _ in range(t):
	a,b, q = [int(i) for i in input().split()]
	a, b = min(a,b), max(a,b)
	jump = int((a*b)/gcd(a,b))
	all_ans = []
	
	reusable = []
	for x in range(jump):
		if (x - x%b)%a != 0:
			reusable.append(1)
		else:
			reusable.append(0)
	sum_reusable = [0]
	for x in range(jump):
		sum_reusable.append(sum_reusable[-1] + reusable[x])

	for _ in range(q):
		l , r = [int(i) for i in input().split()]
		ans = 0
		shift = l%jump
		amt1 = int((r-l+1)/jump)
		# if amt1 >0:
			# ans += sum(reusable[l%jump:])
			# ans += sum(reusable[:r%jump + 1])
		# else:
		if(l%jump > r%jump):
			# print('not here')
			# ans += max(0,amt1)*sum(reusable)
			ans += max(0,amt1)*sum_reusable[-1]
			# ans += sum(reusable[l%jump:]) + sum(reusable[:r%jump + 1])
			ans += (sum_reusable[-1] - sum_reusable[l%jump]) + sum_reusable[r%jump + 1]
		else:
			# print('here')
			# ans += max(0,amt1)*sum(reusable)
			ans += max(0,amt1)*sum_reusable[-1]
			# ans += sum(reusable[l%jump : r%jump + 1])
			ans += sum_reusable[r%jump+1] - sum_reusable[l%jump]
		# for x in range(min(jump, r - l + 1)):
		# 	ans += (int((r-l -(x))/jump) +1)*reusable[(shift + x)%jump]
		all_ans.append(ans)
	print(*all_ans, sep = ' ')

