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>

int main (void)
{
    long long A,B,N,i;
    long long almax=0;
    double alpha;

    scanf("%lld %lld %lld",&A,&B,&N);


    if(A<10000){
        for(i=0;i<N;i++){
                alpha= A*i/B - A*(int )(i/B);

                if(alpha>almax){
                    almax=alpha;
                }
            }
            printf("%lld",(int )almax);
    }else{
        for(i=0;i<N;i++){
                alpha= i/B - (int )(i/B);

                if(alpha>almax){
                    almax=alpha;
                }
            }
            printf("%lld",(int )A*almax);

    }

   



    return 0;
}