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 math import gcd
from sys import stdin
inp = stdin.readline

t = int(inp().strip())


def lcm(a,b):
    return (a*b) / gcd(a,b)


for _ in range(t):
    a, b, q = [int(x) for x in inp().strip().split()]
    period = a*b
    maximum = max(a,b)
    ans = []
    for _2 in range(q):
        l, r = [int(x) for x in inp().strip().split()]
        counter = 0
        ans1 = r-l+1
        while True:
            if a*b*counter <= r:
                lc = lcm(a,b)
                start = 1*lc*counter
                end = 1*lc*counter+maximum
                lowerBound = max(l,start)
                upperBound = min(r+1,end)
                if lowerBound > upperBound:
                    counter += 1
                    continue
                else:
                    ans1 -= upperBound-lowerBound
            elif a*b*counter > r:
                break
            counter += 1
        ans.append(int(ans1))
    print(*ans)