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 pgcd(a,b) :  
   while a%b != 0 : 
      a, b = b, a%b 
   return b
   
    
    
t = int(input())
for _ in range(t):
    c = input().split()
    a = int(c[0])
    b = int(c[1])
    q = int(c[2])
    if a > b:
        b,a = a,b
    p = a//pgcd(a,b)
    out = []
    for _ in range(q):
        lr = input().split()
        l = int(lr[0])
        r = int(lr[1])
        v = r-l+1
        if (l//b)%p == 0:
            v -= min(b-l%b,r-l+1)
        if (r//b) != (l//b) and (r//b)%p == 0:
            v -= r%b+1
        minl = l//b +1
        if minl%p > 0:
            minl += p - minl%p
        maxr = r//b-1
        maxr -= maxr%p
        if maxr >= minl:
            v -= b*((maxr-minl)//p+1)
        out.append(v)
    
                
                
                
        
        
    print(*out,sep=" ")