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:
#author: raghav_0901
from math import *


def _in():
    return [int(x) for x in input().split(' ')]


def main():
    a, b, q = _in()
    lc = a * (b // gcd(a, b))
    arr = [0, ]
    for i in range(1, lc + 1):
        x = int(((i % a) % b) != ((i % b) % a))
        prev = arr[-1]
        arr.append(prev + x)
    # print(arr)
    for i in range(q):
        l, r = _in()
        ans = 0
        p = r // lc
        ans = ans + arr[r - (lc * p)]
        if(l % lc == 0):
            b = (l) // lc
            ans = ans + ((p - b) * arr[-1])
        else:
            b = (l) // lc
            extra = l - (b * lc) - 1
            ans = ans + arr[-1] - arr[extra]
            ans = ans + ((p - b - 1) * arr[-1])
        print(ans, end=" ")
    print()


if __name__ == "__main__":
    tt = 1
    tt = int(input())
    for _ in range(tt):
        main()