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()
{
	double A;
	double B;
	long long N;
	double maxval=0;
	double val;
	long long imax;

	scanf("%lf %lf %lld",&A,&B,&N);
	for(long long i = 1; i<=B;i++)
	{
		val = floor(i*A/B) - A*floor(i/B);
		if (val>maxval)
		{
			maxval=val;
			imax = i;
		}
	}
	for(long long i = imax; i<=N; i+=B)
	{
		val = floor(i*A/B) - A*floor(i/B);
                if (val>maxval)
                {
                        maxval=val;
                }
        }

	printf("%lld\n",(long long)maxval);

	return 0;
}