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:
import math
tll=int(input())
while(tll>0):
	tll-=1
	a,b,q=map(int,input().split())
	while(q>0):
		q-=1
		l,r=map(int,input().split())
		if (a>b):
			a,b=b,a
		if (l<b and r<b):
			print(0, sep=" ")
			continue
		ans=r-l+1
		if (l<b):
			ans-=(b-l)
		lcm=(a*b)//math.gcd(a,b)
		l=max(l,b)
		n1=l//lcm
		n2=r//lcm
		n=n2-n1
		if (l==(n1*lcm)):
			n+=1
		ans-=(n*b)
		if ((n2*lcm)+b-1>r):
			ans+=((n2*lcm)+b-1)-r
		if (((n1-1)*lcm)+b-1>=l):
			ans-=((n1-1)*lcm)+b-l
		if (n1*lcm<l and n1*lcm+b-1>=l):
			ans-=n1*lcm+b-l
		# if (r>lcm):
		# 	n1=r//lcm
		# 	n2=l//lcm
		# 	temp=r%lcm
		# 	if ((n1*lcm)+b-1>=r):
		# 		ans-=(n1-1)*b
		# 		ans-=r-(n1*lcm)+1
		# 	else:
		# 		ans-=(n1)*b
		# 	if (l>lcm):
		# 		ans+=n2*b
		# 		if ((n2*lcm)+b-1>=l):
		# 			ans-=(n2*lcm)+b-l
		print(ans,sep=" ")
	print()