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: Runtime Error)

#include<stdio.h>
#include<math.h>
int main(void){
  int a,b,c,d;
  scanf("%d",&a);
  scanf("%d",&b);
  scanf("%d",&c);
  int e[c];
  for(int x=c;x>0;x--){
    e[c-x]=floor(a*x/b)-a*floor(x/b);
  }
  d=e[0];
  for(int i=0;i<c;i++){
      if(e[i]>=d){
          d-e[i];
      }else{
          d=d;
      }
  }
  printf("%d",d);
  return 0;
}