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: Accepted)

#include <stdio.h>
#include <math.h>
int main(void){
  long long a,b,n,k,i;
  scanf("%lld %lld %lld",&a,&b,&n);
  k= floor((long double)a*n/b);
  if(b==1){
    printf("0");
    } else if(n<b) {
      printf("%lld",k);
    }else{
    k= floor((long double)a*(b-1)/b)-floor((long double)(b-1)/b);
          printf("%lld",k);
  }


  return 0;
}