B. DIV + MOD
math
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 (\(l \le x \le r\)).
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 \(a\).
For each test case, output one number on a separate line — the maximum value of the function on a given segment for a given \(a\).
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.