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 math


def not_equal(a, b, lcm, i):
    if i <= b-1:
        return 0

    # The numbers up to the minimum always satisfy the condition
    out = i - b + 1

    # Subtract the terms k*l + c, k \in N, c < b
    n_seqs = (i - b) // lcm
    out -= n_seqs * b

    # Add the terms that you may have missed when the modulus is close
    mod = i % lcm
    if mod < b:
        out -= (mod+1)

    return out


def naive_not_equal(a, b, val):
    count = 0
    for i in range(val+1):
        if (i % a) % b != (i % b) % a:
            count += 1
    return count


t = int(input())
for _ in range(t):
    a, b, q = map(int, input().split())

    # Put a and b in order (a < b)
    if a > b:
        a, b = b, a

    lcm = a * b // math.gcd(a, b)

    out = []
    for _ in range(q):
        l, r = map(int, input().split())
        lhs = not_equal(a, b, lcm, l-1)
        rhs = not_equal(a, b, lcm, r)
        ans = rhs - lhs
        out.append(str(ans))

    print(' '.join(out))
