Problem: 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 

Solution (in PYTHON3):

Solution:
'''
INPUT SHORTCUTS
N, K = map(int,input().split())
N ,A,B = map(int,input().split())
string = str(input())
arr = list(map(int,input().split()))
N = int(input())
'''

def cont(a,b,dp,x):
	return (dp[(a*b)-1]*(x//(a*b)))+dp[x%(a*b)]


def main():
	a,b,q = map(int,input().split())
	dp = [0 for _ in range(a*b)]
	for i in range(a*b):
		if i==0:
			dp[i]=0
		else:
			lhs = (i%a)%b
			rhs = (i%b)%a
			if lhs!=rhs:
				dp[i]=1
	for i in range(1,a*b):
		dp[i]+=dp[i-1]
	while q:
		l,r = map(int,input().split())
		if a==b:
			print("0",end=' ')
		else:
			# print(dp)
			# print(cont(a,b,dp,r),cont(a,b,dp,l-1))
			print(cont(a,b,dp,r)-cont(a,b,dp,l-1),end=' ')
		q-=1


	
n = int(input())
while n:	
	main()
	# print()
	n-=1