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 gcd(x, y): 
   while(y): 
       x, y = y, x % y 
  
   return x



cases = int(input())
for i in range(cases):
    inp = input().split(" ")
    a = int(inp[0])
    b = int(inp[1])
    q = int(inp[2])
    GCD = gcd(a, b)
    LCM = (a*b)//GCD
    minst = a if a <= b else b
    mest = a if a >= b else b
    result = ""
    for k in range(q):
        inp2 = input().split(" ")
        l = int(inp2[0])
        r = int(inp2[1]) +1 
        innr = (r-r%LCM)
        innl = (l+(LCM - l%LCM))
       
        ans = 0
        if (innr > innl):
            ans = ((innr - innl + 1)//LCM)*(LCM-mest) # alle hele cycles som er i intervallet
        
        if r//LCM == l//LCM: # de er i samme cycle
            tmp = r%LCM - (l%LCM if l%LCM > mest else mest)
            ans += tmp if tmp > 0 else 0
        else:
            tmpl = LCM - (l%LCM if l%LCM > mest else mest)
            tmpr = r%LCM - mest
            ans += tmpl if tmpl > 0 else 0
            ans += tmpr if tmpr > 0 else 0
        
        result += str(ans) + " "
    print(result)