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 PYTHON3):

Solution:
import sys
import math

t = int(input())

for _ in range(t):
    a, b, q = [int(c) for c in input().split()]
    a, b = min(a, b), max(a, b)
    pgcd = math.gcd(a, b)
    ppcm = a*b//pgcd

    cumulative_count = [0 for i in range(ppcm)]
    count = 0
    for i in range(ppcm):
        if i % a % b != i % b % a:
            count += 1
        cumulative_count[i] = count

    cumul_per_period = cumulative_count[-1]
    for i in range(q):
        li, ri = [int(c) for c in input().split()]

        count_li = cumul_per_period * ((li-1) // ppcm) + cumulative_count[(li-1) % ppcm]
        count_ri = cumul_per_period * (ri // ppcm) + cumulative_count[ri % ppcm]
        count = count_ri - count_li
        
        print(count, end=' ')
    print('')