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: Accepted)

/*
maximize: floor(Ax/B) - A * floor(x/B)
maximize: floor(Ax/B - A * floor(x / B))
floor前を最大化しても解は変化しない
maximize: Ax/B - A * floor(x / B)
maximize: A(x/B -  floor(x / B))
maximize: x/B -  floor(x / B)
 */

use std::io::Read;

fn main() {
    let mut line = String::new();
    std::io::stdin().read_to_string(&mut line);
    let mut iter = line.split_whitespace();
    let a: i64 = iter.next().unwrap().parse().unwrap();
    let b: i64 = iter.next().unwrap().parse().unwrap();
    let n: i64 = iter.next().unwrap().parse().unwrap();
    if b <= n {
        let x = b - 1;
        println!("{}", a * x / b - a * (x / b))
    } else {
        let x = n;
        println!("{}", a * x / b - a * (x / b))
    }

}