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)

use proconio::{
   fastout, input,
   marker::{Bytes, Chars, Isize1, Usize1},
};

use num::integer::*;

fn f(a: isize, b: isize, i: isize) -> isize {
   return ((a * i) as f64 / b as f64) as isize - a * (i as f64 / b as f64) as isize;
}

fn main() {
   input! {
      a:isize,b:isize,n:isize
   }
   let mut ans = 0_isize;
   ans = std::cmp::max(ans, f(a, b, n));
   let e = lcm(a, b) / a;
   let g = lcm(a, b) / b;
   for i in -100..=100 {
      if 0 <= (n / b) * b - i && (n / b) * b - i <= n {
         ans = std::cmp::max(ans, f(a, b, (n / b) * b - i));
      }
      if 0 <= (n / e) * e - i && (n / e) * e - i <= n {
         ans = std::cmp::max(ans, f(a, b, (n / e) * e - i));
      }
      if 0 <= (n / g) * g - i && (n / g) * g - i <= n {
         ans = std::cmp::max(ans, f(a, b, (n / g) * g - i));
      }
   }
   for i in 0..=std::cmp::min(1000000, n) {
      ans = std::cmp::max(ans, f(a, b, i));
   }

   println!("{}", ans);
}