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 PYTHON):

Solution:
from sys import stdin, stdout
T = int(stdin.readline())


def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def lcm(a, b):
    return (a / gcd(a, b))*b


for p in xrange(T):
    a, b, q = map(int, stdin.readline().split())
    if a > b:
        a, b = b, a
    L = lcm(a, b)
    resL = resR = 0
    for m in xrange(q):
        l, r = map(int, stdin.readline().split())
        l -= 1
        if L == b:
            stdout.write('0 ')
            continue
        if r/L > 0:
            resR = (r/L)*(L-b)
            resR += max(0, (r-((r/L)*L+b))+1)
        else:
            resR = max(0, r-b+1)

        if l/L > 0:
            resL = (l/L)*(L-b)
            resL += max(0, (l-((l/L)*L+b))+1)
        else:
            resL = max(0, l-b+1)

        stdout.write(str(resR-resL)+' ')
    stdout.write('\n')