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:
n_cases = int(input())
for case_num in range(n_cases):
    a, b, q = [int(x) for x in input().split()]
    k = a*b
    pref = [0] * k
    for i in range(1, k):
        pref[i] = pref[i-1]
        if (i % a) % b != (i % b) % a:
            pref[i] += 1
    for i, query in enumerate(range(q)):
        l, r = [int(x) for x in input().split()]
        count = 0
        if l % k != 0:
            count += pref[k-1] - pref[(l % k) - 1]
            l = ((l // k) + 1) * k
        if r % k != k-1:
            count += pref[(r % k)]
            r = (r // k) * k - 1
        assert (r - l + 1) % k == 0
        count += pref[k-1] * ((r - l + 1) // k)
        if i > 0:
            print(' ', end='')
        print(count, end='')
    print()