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:
def lcm(x, y):  
   if x > y:  
       greater = x  
   else:  
       greater = y  
   while(True):  
       if((greater % x == 0) and (greater % y == 0)):  
           lcm = greater  
           break  
       greater += 1  
   return lcm  
 
t=int(input())
for _ in range(t):
  a,b,q=map(int,(input().split()))
  x=lcm(a,b)
  for i in range(q):
    count=0
    
    l,r=map(int,(input().split()))
    lp=(l//(x)+1)*x
    rp=(r//(x))*x
    count+=((rp-lp)//(x)+1)*max(a,b)
    #print([rp,lp])
    if l%(x)<max(a,b):
      count+=max(a,b)-l%(x)
    if r%(x)<max(a,b):
      count-=max(a,b)-r%(x)-1
    print(int(r-l-count+1),end=" ")
    #print(int(count))
  print()