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 

'''
a, b = 7, 10
print a, b
for x in range(100) :
  print x, ((x % a) % b), '!=', ((x % b) % a), " : ", ((x % a) % b) != ((x % b) % a)
'''

def gcd(a, b) :
  while b :
    a, b = b, a % b
  return a

T = input()
for t in range(T) :
  A, B, Q = map(int,raw_input().split())
  period = min(A, B) // gcd(A, B) * max(A, B)

  for q in range(Q) :
    l, r = map(int,raw_input().split())
  
    if max(A,B) % min(A,B) == 0 : print 0
    else :
      lp = l // period
      rp = r // period

      begin, middle, end = 0, 0, 0
      
      if lp < rp : 
        nxt = (lp+1)*period
        begin = min(period-max(A,B), nxt - l)
        rb = r-rp*period
        end = max(0, rb - max(A, B)+1)
      elif lp == rp :
        lb = max(l-lp*period, max(A,B))
        rb = r-rp*period
        begin = max(0, rb - lb + 1)

      if lp - rp > 1 : 
        middle = (rp - (lp+1)) * (period-max(A, B))

      print begin + middle + end,
  
  print ""
