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 
import os
import sys


def calc_ts(a, b):
    r = [0] * (a * b)
    s = 0
    for x in range(a * b):
        s += (x % a % b) != (x % b % a)
        r[x] = s

    return r

def f(a, b, l, r, ts):
    nr, mr = divmod(r, a * b)
    nl, ml = divmod(l - 1, a * b)
    return ts[-1] * nr + ts[mr] - (ts[-1] * nl + ts[ml])



def bf(a, b, l, r):
    return sum((x % a % b) != (x % b % a) for x in range(l, r + 1))


if False:
    import numpy as np

    for i in range(100):
        a, b = np.random.choice(100, 2) + 1
        ts = calc_ts(a, b)
        l, r = sorted(np.random.choice(100000, 2, replace=False) + 1)
        assert f(a, b, l, r, ts) == bf(a, b, l, r)


def pp(input):
    n_test = int(input())
    for i in range(n_test):
        a, b, q = map(int, input().split())
        res = [0] * q
        ts = calc_ts(a, b)
        for idx in range(q):
            l, r = map(int, input().split())
            res[idx] = f(a, b, l, r, ts)

        print(" ".join(map(str, res)))


if "paalto" in os.getcwd():
    from string_source import string_source

    pp(
        string_source(
            """2
4 6 5
1 3
1 1
1 5
1 7
1 9
7 10 2
7 8
100 200"""
        )
    )


else:
    pp(iter(sys.stdin.buffer.read().decode().splitlines()).__next__)
