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: Time Limit Exceeded)

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);
 
impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> IO<R, W> {
        IO(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: std::ops::Deref<Target = str>>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}

fn solve(a: f64, b: f64, n: i64) -> i32 {
    let mut max_val: f64 = 0.0;
    for x in 0..=n {
        let x = x as f64;
        let ans = f64::floor(a*x/b) - a * f64::floor(x/b);
        max_val = max_val.max(ans);
    }
    max_val as i32
}

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    let ans = solve(sc.read(), sc.read(), sc.read());
    println!("{}", ans);
}