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()
{
  long long a,b,n,i,max,f1,f2,sum,k;
  k=1;
  max=0;
  scanf ("%lld",&a);
  scanf ("%lld",&b);
  scanf ("%lld",&n);
  for (;n;--n)
    {
      f1=floor((a*k)/b);
      f2=floor(k/b);
      sum = f1-a*f2;
      if (sum > max)
        max=sum;
      k++;
    }
  printf ("%lld\n",max);
  return 0;
}