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)

extern crate proconio;
use proconio::input;
 
fn main() {
  input! {
    a: i64,
    b: i64,
    n: i64
  }
  
  let mut x = n;
  
  if(a%b == 1) {
    println!("{}", n);
    return;
  }
  
  if n % b == b - 1 {
    println!("{}", (((a%b)*(x%b))%b - x%b));
  }
  
  let mut last = ((a%b)*(x%b))%b;
  while(x > 0) {
    x = x - 1;
    let now = ((a%b)*(x%b))%b;
    if now <= last {
      x = x + 1;
      println!("{}", (((a%b)*(x%b))%b - x%b));
      return;
    }
  }
  
  x = n;
  println!("{}",  (((a%b)*(x%b))%b - x%b));
}