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 functools
import math
import sys

# sys.stdin = open('c.in')
t = int(sys.stdin.readline())


def solve(a, b, queries):
    if a > b:
        a, b = b, a
    gcd = math.gcd(a, b)
    lcm = gcd * (a // gcd) * (b // gcd)
    result = []

    @functools.lru_cache(None)
    def count_to_x(x):
        if x < lcm:
            return 0 if x < b else x - b + 1
        else:
            r = (x // lcm) * b + min(x - (x // lcm * lcm) + 1, b)
        return (x + 1) - r

    for li, ri in queries:
        result.append(count_to_x(ri) - count_to_x(li - 1))
    return result


for _ in range(t):
    a, b, q = map(int, sys.stdin.readline().split())
    queries = []
    for _ in range(q):
        queries.append(list(map(int, sys.stdin.readline().split())))
    r = solve(a, b, queries)
    print(' '.join(map(str, r)))
