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 (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

int main(void) {
  double a,b,n;
  scanf("%lf %lf %lf",&a,&b,&n);
  unsigned long int max;
  max=0;
  for(double x=0;x<=n;x++){
    if(x/b-(int)(x/b)<0.5) continue;
    if(max<((int)(a*x/b)-a*(int)(x/b))) max=(int)(a*x/b)-a*(int)(x/b);
  }
  printf("%ld",max);
  return 0;
}