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 math
import os,io
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def findv(lcm,l,r,b):
    p = max(min(r,b),l)
    s = r-p+1
    x1 = p//lcm
    x2 = r//lcm
    if x1*lcm+b > p:
        s -= b-p%lcm
    x1 += 1
    s -= b*(x2-x1+1) if (x2*lcm+b <= r) else (b*(x2-x1)+r%lcm+1)
    # if x2*lcm+b <= r:
    #     s -= b*(x2-x1+1)
    # else:
    #     s -= b*(x2-x1)+ r%lcm + 1
    return s

cases = int(input())
for t in range(cases):
    a,b,q = list(map(int,input().split()))
    a,b = min(a,b),max(a,b)
    a = (a*b)//math.gcd(a,b)
    out = []
    for i in range(q):
        l,r = list(map(int,input().split()))
        if b>r:
            out.append(0)
        else:
            out.append(findv(a, l, r, b))
    print(*out)