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 
from sys import stdin
from math import gcd

def func(a, b, l, r):
    if r < b or a == 1 or b == 1:
        return 0
    l = max(l, b)
    total = r-l+1
    if total <= b:
        count = 0
        for i in range(l, r+1):
            if (i%a)%b != (i%b)%a:
                count += 1
        return count
    M, m = r//lcm, l//lcm
    count = M-m
    if M == 0:
        return total
    ans = min(r-(M*lcm)+1, b)
    count -= 1
    if m > 0:
        ans += min(b, ((m+1)*lcm)-l+1)
        count -= 1
    return total - ans - count*b
for _ in range(int(stdin.readline())):
    #n = int(stdin.readline())
    #l = list(map(int, stdin.readline().split()))
    w = list(map(int, stdin.readline().split()))
    #arr = list(map(int, stdin.readline().split()))
    a, b = min(w[0], w[1]), max(w[0], w[1])
    lcm = (a*b)//gcd(a, b)
    ans = []
    for i in range(w[2]):
        l, r = map(int, stdin.readline().split())
        if lcm == b:
            ans.append(0)
        else:
            ans.append(func(a, b, l, r))
    print(*ans)
    
