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

    count = 0
    cum = []
    for i in range(0, (a*b)): # cycle of ab
        if ((i % a) % b) != ((i % b) % a):
            count += 1
        cum.append(count)

    one_cycle_total = cum[a*b-1]

    for _ in range(q):
        l, r = map(int, input().split())

        # solve from 0 to r (imagine l=0)
        complete_cycles = r // (a*b)
        extra_right_part = r % (a*b)
        answer = one_cycle_total * complete_cycles + cum[extra_right_part]

        # solve from 0 to l-1
        complete_cycles = (l-1) // (a*b)
        extra_right_part = (l-1) % (a*b)
        answer2 = one_cycle_total * complete_cycles + cum[extra_right_part]

        print(answer - answer2)

        # 
        # # cycles of ab
        # biggest = 0
        # for i in range(l, r+1):
        #     if i % (a*b) == 0:
        #         biggest = i # largest multiple of ab
        # smallest = 0
        # for i in range(l, r+1):
        #     if i % (a*b) == 0:
        #         smallest = i
        #         break # smallest multiple of ab
        # len = (biggest - 1 - smallest) // (a * b) + 1 # number of ab lengths
        # total = len * count # number of x that satisfy in ab lengths
        #
        # # numbers not in the cycles of ab
        # for i in range(l, smallest):
        #     if (i % (a*b))
        #         total += 1 # front
        # for i in range(biggest, r+1):
        #     if
        #         total += 1 # back
        #
        # print(total)