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 
from sys import stdin, stdout

T = int(stdin.readline())

for _ in range(T):
    [a, b, q] = [int(x) for x in stdin.readline().split()]
    maxAB = max(a, b)
    aGCD = a
    bGCD = b
    while aGCD != 0 and bGCD != 0:
        if aGCD > bGCD:
            aGCD %= bGCD
        else:
            bGCD %= aGCD

    gcd = aGCD + bGCD

    nok = a * b // gcd

    batch = nok - maxAB
    resArr = []

    for __ in range(q):
        [l, r] = [int(x) for x in stdin.readline().split()]
        res = 0
        if a == 1 and b == 1:
            res = 0
        else:
            res += ((r - l) // nok) * batch

            if r % nok >= l % nok:
                if r % nok >= maxAB:
                    if l % nok >= maxAB:
                        res += r % nok - l % nok + 1
                    else:
                        res += r % nok - maxAB + 1
            else:
                if r % nok >= maxAB:
                    res += r % nok - maxAB + 1
                res += min(batch, nok - l % nok)
        resArr.append(str(res))

    stdout.write(' '.join(resArr) + "\n")
