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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>

long long	calc(long long a, long long b, unsigned long long x)
{
  	long long	c;
  	long long 	d;
  	
  	c = floor(a * x / b);
  	d = a * floor(x / b);
  	return (c - d);
}

int	main(void)
{
  	long long	a;
  	long long 	b;
  	unsigned long long 	n;
  	unsigned long long 	x;
  	long long 	max;
  	long long 	tmp;
  	  
  	scanf("%lld %lld %lld", &a, &b, &n);
  
  	max = -900000000000;
  	if (n < b)
  	{
      	for(x=0; x<=n; x++)
        {
          	tmp = calc(a, b, x);
          	if (max < tmp)
              	max = tmp;
        }
    }
    else
    {
      	for(x=0; x<b; x++)
        {
          	tmp = calc(a, b, x);
          	if (max < tmp)
              	max = tmp;
        }
      	for(x=b; x<=n; x+=b)
        {
          	tmp = calc(a, b, x);
          	if (max < tmp)
              	max = tmp;
        }
    }
  	printf("%lld", max);
  	return (0);
}