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 
from sys import stdin,stdout

def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a) 
  
# Function to return LCM of two numbers 
def lcm(a,b): 
    return (a*b) / gcd(a,b)

t=int(input())

for i in range(t):

	a,b,q=list(map(int,stdin.readline().split()))

	d=(int(lcm(a,b)))

	list1=[]

	maxi=max(a,b)

	# print(d)

	# t=int(input())

	for i in range(q):

		l,r=list(map(int,stdin.readline().split()))

		if a==b:

			# print(0)

			list1.append(0)


		elif r<=maxi:


			cnt=0

			for i in range(l,r+1):

				# print((i%a)%b,end=" ")

				# print((i%b)%a)

				if (i%a)%b==(i%b)%a:
					cnt+=1

			list1.append(r-l+1-cnt)

		else:



			rem1=l%d

			quo1=l//d

			rem2=r%d

			quo2=r//d
			sum1=0

			if rem1<=maxi:

				sum1=sum1+maxi-rem1

				if rem2<=maxi:

					sum1+=(quo2-quo1-1)*(maxi)
					sum1+=rem2+1

				else:

					sum1+=(quo2-quo1)*(maxi)


			else:

				if rem2<=maxi:

					sum1+=(quo2-quo1-1)*(maxi)
					sum1+=rem2+1

				else:

					sum1+=(quo2-quo1)*(maxi)

			# print(sum1)


			sum2=r-l+1

			list1.append(sum2-sum1)


	for i in list1:

		print(i,end=" ")

	print()













		


