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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <cstdint>
#include <cstdio>
#include <algorithm>

int main() {
  intmax_t a, b, n;
  scanf("%jd %jd %jd", &a, &b, &n);

  auto f = [&](intmax_t x) {
    return (a*x) / b - a * (x/b);
  };

  intmax_t res = 0;
  for (intmax_t i = 1; i*i <= n; ++i) {
    res = std::max({res, f(i), f(n/i)});
  }

  printf("%jd\n", res);
}