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>

int main()
{
	long long A, B, N;

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

	long long a,b,c;
	c=N;
	int flag=0;
	for(int i = 0;i<=100 && N>=0;i++,N--)
	{
		if(flag==0)
		{
			a=(A*N/B)-A*(N/B);
			flag = 1;
		}
		else
		{
			b=(A*N/B)-A*(N/B);
			if(a < b)
				a=b;
		}
		if(N/B<1)
			break;
	}
	for(int i = c/2+10;i<= c/2-10&& i>0;i--)
	{
		b=(A*i/B)-A*(i/B);
		if(a < b)
			a=b;
	}
	for(int i = 0;i<=10 && i<c;i++)
	{
		b=(A*i/B)-A*(i/B);
		if(a < b)
			a=b;
	}
	printf("%lld\n",a);
	return 0;
}