Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

macro_rules! line {
    () => {{
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf
    }};
}
macro_rules! line_as_vec {
    ( $t:ty ) => {
        ln!()
            .trim()
            .split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect::<Vec<$t>>()
    };
}
macro_rules! input {
    { $( $i:ident : $t:ty ),* } => {
        let ( $( $i ),* ) = match &line!().split_whitespace().collect::<Vec<_>>()[..] {
            &[ $( $i ),* ] => ( $( $i.parse::<$t>().unwrap() ),* ),
            _ => panic!(),
        };
    };
}
fn main() {
    input! {a: f64, b: f64, n: f64};

    println!(
        "{}",
        if n > b {
            a as u64 - 1
        } else {
            (n * (a / b)) as u64
        }
    );
}