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 get_shift_freq(M, m):
    k = 1
    while 1:
        if (k * (M % m)) % m == 0:
            return k
        k += 1

for testcase in range(int(input())):
    a, b, q = map(int, input().split())
    M = max(a, b)
    m = min(a, b)
    shift = M % m
    shift_freq = get_shift_freq(M, m)
    shift_freq_i = shift_freq * M
    answers = []
    for ii in range(q):
        ans = 0
        l, r = map(int, input().split())
        if (l % shift_freq_i) < M:
            pass
        else:
            ans -= (l % shift_freq_i)
            ans += M
        l -= l % shift_freq_i
        rpos = r % shift_freq_i
        if rpos < M:
            ans -= M * (shift_freq - 1)
        else:
            ans -= shift_freq_i - rpos - 1
        r += shift_freq_i - rpos
        ans += ((r - l) // shift_freq_i) * ((shift_freq - 1) * M)
        answers.append(ans)
    print(*answers)