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 count(f,i,n):
    #print(f'f[-1]*{i//n}+f[{i%n}]='+str(f[-1]*(i//n)+f[i%n]))
    return f[-1]*(i//n)+f[i%n]

for _ in range(int(input())):
    a,b,q=[int(x)for x in input().split()]
    f=[0]
    n=a*b
    for i in range(1,n):
        f.append(f[i-1])
        if (i%a)%b != (i%b)%a:
            f[i] = f[i] + 1
    #print(f)
    res = []
    for _ in range(q):
        l, r = [int(x) for x in input().split()]
        res.append(count(f,r,n)-count(f,l-1,n))

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