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));
   if n < 10000 {
      for i in 0..=n {
         ans = std::cmp::max(ans, f(a, b, i));
      }
   } else {
      let mut q = b - 1;
      let mut e = lcm(a, b) / b;
      let mut v = e;
      let mut w = (n / e) * e;
      let mut c = 0;
      while true {
         if q <= n {
            ans = std::cmp::max(ans, f(a, b, q));
            ans = std::cmp::max(ans, f(a, b, w));
         }
         if v <= n {
            ans = std::cmp::max(ans, f(a, b, v - 1));
         }
         if w > 0 {
            ans = std::cmp::max(ans, f(a, b, w - 1));
         }
         q += b;
         w -= b;
         c += 1;
         v += e;
         w -= e;
         if c > 10000000 {
            break;
         }
      }
   }
   println!("{}", ans);
}