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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>

int main(void)
{
  long int i, a, b, n, max = 0, fa, fb;
  
  scanf("%ld %ld %ld", &a, &b, &n);
  
  if( n > b ){
      n = b;
  }
  
  for( i = 1; i <= n; i++ ){
      fa = a*i/b;
      fb = i/b;
      if( max < fa - (a*fb) ){
          max = fa - (a*fb);
      }
  }
  

  printf("%ld\n", max);
  return 0;
}