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:
# This sequence is periodic every A*B queries.

t = int(input())

for case in range(1, t+1):
    a, b, q = list(map(int, input().split()))
    # generate cases
    res = [((x % a) % b) != ((x % b) % a) for x in range(a*b)]
    running_total = [0 for x in range(a*b)]
    running_total[0] = 0
    for x in range(1, a*b):
        running_total[x] = running_total[x-1] + int(res[x])

    totals = []
    for _ in range(q):
        l, r = list(map(int, input().split()))
        total = 0
        total += (r // (a*b)) * running_total[-1]
        total -= ((l-1) // (a*b)) * running_total[-1]
        total += running_total[r % (a*b)]
        total -= running_total[(l+a*b-1) % (a*b)]
        totals.append(str(total))
    print(' '.join(totals))