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 
def compute_lcm(x, y):
 
   # choose the greater number
   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

def diffsum(x, r):
    return max(0, x-r+1)

t = int(input())
for tt in range(t):
    a, b, q = map(int, input().split())
    a, b = min(a, b), max(a, b)

    lcm = compute_lcm(a, b)
    lcmsum = lcm-b
 
    sol = []
    for qq in range(q):
        l, r = map(int, input().split())
        lm = (l-1) % lcm
        sol.append(-diffsum(lm, b))
 
        rm = (r % lcm)
        sol[-1] -= lcmsum - diffsum(rm, b)#sum(c[-rm:])
 
        sol[-1] += (r+lcm-rm - l+lm+1)//lcm * lcmsum
 
    print(' '.join(map(str, sol)))