DIV + MOD

2000ms

262144K

Description:

Not so long ago, Vlad came up with an interesting function:

$$$f_a(x)=\left\lfloor\frac{x}{a}\right\rfloor + x \bmod a$$$, where $$$\left\lfloor\frac{x}{a}\right\rfloor$$$ is $$$\frac{x}{a}$$$, rounded down, $$$x \bmod a$$$ — the remainder of the integer division of $$$x$$$ by $$$a$$$.

For example, with $$$a=3$$$ and $$$x=11$$$, the value $$$f_3(11) = \left\lfloor\frac{11}{3}\right\rfloor + 11 \bmod 3 = 3 + 2 = 5$$$.

The number $$$a$$$ is fixed and known to Vlad. Help Vlad find the maximum value of $$$f_a(x)$$$ if $$$x$$$ can take any integer value from $$$l$$$ to $$$r$$$ inclusive ($$$l \le x \le r$$$).

Input:

The first line of input data contains an integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of input test cases.

This is followed by $$$t$$$ lines, each of which contains three integers $$$l_i$$$, $$$r_i$$$ and $$$a_i$$$ ($$$1 \le l_i \le r_i \le 10^9, 1 \le a_i \le 10^9$$$) — the left and right boundaries of the segment and the fixed value of $$$a$$$.

Output:

For each test case, output one number on a separate line — the maximum value of the function on a given segment for a given $$$a$$$.

Sample Input:

5
1 4 3
5 8 4
6 10 6
1 1000000000 1000000000
10 12 8

Sample Output:

Note:

In the first sample:

$$$f_3(1) = \left\lfloor\frac{1}{3}\right\rfloor + 1 \bmod 3 = 0 + 1 = 1$$$,
$$$f_3(2) = \left\lfloor\frac{2}{3}\right\rfloor + 2 \bmod 3 = 0 + 2 = 2$$$,
$$$f_3(3) = \left\lfloor\frac{3}{3}\right\rfloor + 3 \bmod 3 = 1 + 0 = 1$$$,
$$$f_3(4) = \left\lfloor\frac{4}{3}\right\rfloor + 4 \bmod 3 = 1 + 1 = 2$$$

As an answer, obviously, $$$f_3(2)$$$ and $$$f_3(4)$$$ are suitable.

Informação

Provedor Codeforces

Código CF1650B