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 
import sys
if sys.subversion[0] == "PyPy":
    import io, atexit
    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
    
    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()
 
RS = raw_input
RI = lambda x=int: map(x,RS().split())
RN = lambda x=int: x(RS())
''' ...................................................................... '''

def lcm(a,b):
    def gcd(a,b):
        while b:
            a,b = b,a%b
        return a

    return a*b/gcd(a,b)

def count(L,freq,per):
    same = (L/per)*freq[per]
    same += freq[L%per]
    return L-same

for _ in xrange(RN()):
    a,b,q = RI()
    per = lcm(a,b)

    freq = [0]*(per+1)
    #freq[0] = 1
    for i in xrange(1,per+1):
        freq[i] = freq[i-1]
        if (i%a)%b == (i%b)%a:
            freq[i]+=1

    out = []
    for Q in xrange(q):
        l,r = RI()
        ans = count(r,freq,per)-count(l-1,freq,per)
        out.append(ans)
    print ' '.join(map(str,out))
        
