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 
t=int(input())
for you in range(t):
	l=input().split()
	a=int(l[0])
	b=int(l[1])
	q=int(l[2])
	arr=[0 for i in range(a*b)]
	for i in range(a*b):
		if(((i%b)%a)!=((i%a)%b)):
			arr[i]=1
	prefsum=[0 for i in range(a*b)]
	sumnow=0
	for i in range(a*b):
		sumnow+=arr[i]
		prefsum[i]=sumnow
	full=prefsum[-1]
	arr=[0 for i in range(q)]
	for i in range(q):
		ans=0
		l=input().split()
		L=int(l[0])
		R=int(l[1])
		ansl=full*((L-1)//(a*b))
		x=(L-1)%(a*b)
		ansl=ansl+prefsum[x]-prefsum[0]
		ansr=full*((R)//(a*b))
		x=(R)%(a*b)
		ansr=ansr+prefsum[x]-prefsum[0]
		arr[i]=ansr-ansl
	for i in arr:
		print(i,end=" ")
	print()