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
input=sys.stdin.readline

test=int(input())

for t in range(test):
	a,b,q=map(int,input().split())
	m=(a*b)

	s=0
	narr=[0]*m
	for j in range(m):
		if (j%a)%b!=((j%b)%a):
			s+=1
		narr[j]=s
	for i in range(q):		
		l,r=map(int,input().split())
		cnt=0
		cnt=(r // m) * s + narr[r % m]-(((l-1) // m) * s + narr[(l-1) % m])
		# ncnt=0
		# if arr[l%m]==-1:
		# 	for j in range(l,l+m):
		# 		if (j%a)%b!=((j%b)%a):
		# 			cnt+=1
		# 	arr[l%m]=cnt
		# else:
		# 	cnt=arr[l%m]
		# cnt=cnt*fin
		# cnt+=ncnt
		# for j in range(r,r-lef-1,-1):
		# 	if (j%a)%b!=((j%b)%a):
		# 		cnt+=1
		print(cnt,end=" ")
	print()
		# fin=0

		# if l+m<=r:
		# 	fin=r+1
		# else:
		# 	fin=l+m
		# for j in range(l,fin):

		# 	if (j%a)%b!=((j%b)%a):
		# 		cnt+=1
		# if fin!=l+m:
		# 	cnt=cnt*((r-l)//m)
		# 	for j in range(r,(r-l)%m):
		# 		if (j%a)%b!=((j%b)%a):
		# 			cnt+=1
		# print(cnt)