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:
# list(map(int, input().split()))
def nok(a, b):
    x = a
    while x % b != 0:
        x += a
    return x

rw = int(input())
for weqeq in range(rw):
    a, b, q = map(int, input().split())
    lr = [[] for i in range(q)]
    for i in range(q):
        x, y = map(int, input().split())
        lr[i].append(x)
        lr[i].append(y)
    abmod = [0]
    bamod = [0]
    c = nok(a, b)
    pref = [0] * c
    for i in range(1, c):
        abmod.append((i % a) % b)
        bamod.append((i % b) % a)
        if abmod[i] != bamod[i]:
            pref[i] = pref[i - 1] + 1
        else:
            pref[i] = pref[i - 1]
    for i in lr:
        x = (i[0] - 1) % c
        y = i[1] % c
        xp = pref[x] + pref[c - 1] * ((i[0] - 1) // c)
        yp = pref[y] + pref[c - 1] * (i[1] // c)
        n = yp - xp
        print(n, end = ' ')
    print('')