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:
def hcf(a, b):  
    if a == 0 : 
        return b        
    return hcf(b%a, a)
for t in range(int(input())):
    a,b,q1=map(int,input().split())
    if b<a:
        a,b=b,a
    lcm=(a*b)//hcf(a,b)
    for i in range(q1):
        l,r=map(int,input().split())
        c1=r//lcm
        l=l-1
        if l==0:
            c2=0
        else:
            c2=l//lcm
        if c1==0:
            p=0
        else:
            p=(c1-1)*b
        if r>=b:
            p=p+b-1
        else:
            p=p+r
        if r-(c1*lcm)>=b and c1>0:
            p=p+b
        else:
            if c1>0:
                p=p+(r-(c1*lcm)+1)
        if c2>0:
            q=(c2-1)*b
        else:
            q=0
        if l>=b:
            q=q+b-1
        else:
            q=q+l
        if l-(c2*lcm)>=b and c2>0:
            q=q+b
        else:
            if c2>0:
                q=q+(l-(c2*lcm)+1)
        ans=r-l
        ans=ans-(p-q)
        print(ans)