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)

#![allow(non_snake_case)]
#![allow(unused_imports)]
 
use proconio::{input, fastout};
use proconio::marker::*;
use whiteread::parse_line;
use std::collections::*;

#[fastout]
fn main() {
    input!{
        a: usize,
        b: f64,
        n: usize,
    }
    let mut ans_vec: Vec<usize> = vec![0];
    for (i, x) in (1..=n).enumerate() {
        if i == 0 { continue }
        let tmp = ((a * x) as f64 / b) as usize - a * (x as f64 / b) as usize;
        if tmp != 0 { ans_vec.push(tmp) } else { break }
    }

    println!("{}", ans_vec.iter().max().unwrap());
}