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 math

def lcm(x, y):
    return x * y // math.gcd(x, y)

def prefix(l, a, b):
    if l == 0:
        return 0
    res = max(l - max(a, b) + 1, 0)
    num_period = l // lcm(a, b) 
    if num_period > 0 and lcm(a, b) * num_period + max(a, b) - 1 > l:
        res -= l - (lcm(a, b) * num_period) + 1
        res -= max(num_period - 1, 0) * max(a, b)
    else:
        res -= max(num_period, 0) * max(a, b)
    return res

def brute_force(l, r, a, b):
    res = 0
    for i in range(l, r+1):
        if (i%a)%b != (i%b)%a:
            print(i)
            res += 1
    return res

#print(prefix(9, 4, 6))
#print(brute_force(1, 15, 2, 5))

t = int(input())
for _ in range(t):
    a, b, q = [int(x) for x in input().split()]
    for _ in range(q):
        l, r = [int(x) for x in input().split()]
        print(prefix(r, a, b) - prefix(l-1, a, b) , end =' ')
    print()