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 
#!/usr/bin/env python3
import os
from sys import stdin


def gcd(a, b):
    while a > 0:
        b, a = a, b % a
    return b


def solve(tc):
    a, b, q = map(int, stdin.readline().split())
    if a > b:
        a, b = b, a

    g = a*b//gcd(a, b)
    ans = []
    for i in range(q):
        l, r = map(int, stdin.readline().split())

        group = r // g
        if group > 0:
            cnt = group * b + min(r % g, b-1)
        else:
            cnt = min(r % g, b-1)
        res = r - cnt

        l -= 1
        group = l // g
        if group > 0:
            cnt = group * b + min(l % g, b-1)
        else:
            cnt = min(l % g, b-1)
        res -= l - cnt
        ans.append(res)

    print(' '.join([str(x) for x in ans]))


tcs = int(stdin.readline())

tc = 1
while tc <= tcs:
    solve(tc)
    tc += 1
