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 
from sys import stdin, setrecursionlimit, stdout
#setrecursionlimit(1000000)
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def lsi():
    x=list(stdin.readline())
    x.pop()
    return x
def si(): return stdin.readline().rstrip()
#mod=1000000007
res=['YES', 'NO']
############# CODE STARTS HERE #############
test_case=ii()
while test_case:
    test_case-=1



    ans=[]
    a, b, q=mi()
    m=a*b//gcd(a, b)
    if a>b:
        a, b=b, a
    #print(m, a, b)
    for _ in range(q):
        l, r=mi()
        x=l//m
        y=r//m
        l=max(l, x*m+b)
        if r<y*m+b:
            r=y*m-1
            y-=1
        cnt=max(0, r-l+1)
        if cnt:
            cnt-=b*(y-x)
        #print(l, r)
        ans.append(cnt)
    print(*ans)