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 
def lcm(x, y):
    if x > y:
       greater = x
    else:
       greater = y
    while(True):
        if((greater % x == 0) and (greater % y == 0)):
            lcm = greater
            break
        greater += 1
    return lcm

T = int(input())

for tc in range(T):
    a, b, q = map(int, input().split())
    res = []
    maxx = max(a, b)
    mf = lcm(a, b)

    for i in range(q):
        l, r = map(int, input().split())
        count = 0
        counte = 0
        if maxx > r:
            count = 0
        elif maxx == r:
            count = 1
        elif maxx > l:
            counte = maxx - l
            counte += (r//mf - 1) * maxx
            counte += min(maxx, r - (r//mf)*mf + 1)
            count = (r - l + 1) - counte
        else:
            counte += (r//mf - 1) * maxx - (l//mf) * maxx + max(0, ((l//mf)*mf+maxx - l))
            counte += min(maxx, r - (r//mf)*mf + 1)
            count = (r-l+1) - counte





        res.append(str(count))

    print(" ".join(res))




