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<stdio.h>
#include<math.h>
int main()
{
   int a,b,n,x,s;
   scanf("%d%d%d",&a,&b,&n);
   if(a>=b&&b>=n)
   {
       x=n;
       s=(a*x)/b;
   }
   else if(a>=b&&b<n)
   {
       x=n;
       s=(a*x)/b -a*(x/b);
   }
    else if(a<b&&b<n)
   {
       x=n;
        s=(a*x)/b -a*(x/b);
   }
  else  if(a<b&&b>n)
   {
       x=n;
       s=(a*x)/b;
   }

   printf("%d",s);
}