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.buffer.readline
def I(): return(list(map(int,input().split())))
def sieve(n):
	a=[1]*n
	for i in range(2,n):
	    if a[i]:
	        for j in range(i*i,n,i):
	            a[j]=0
	return a

for __ in range(int(input())):
	a,b,q=I()
	pref=[0]*(a*b)
	pref[0]=1

	for i in range(1,a*b):
		if i%a%b==i%b%a:
			pref[i]=pref[i-1]+1
		else:
			pref[i]=pref[i-1]


	for lol in range(q):
		l,r=I()
		l-=1
		tillL=pref[-1]*(l//(a*b))+pref[l%(a*b)]
		tillR=pref[-1]*(r//(a*b))+pref[r%(a*b)]
		print(r-l-(tillR-tillL),end=" ")
	print()


	



