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<string.h>
long long int floor(long long int p,long long int q)
{
	return p / q;
}
int main()
{
	long long int a, b, n,i,max=0,num=0;
	scanf("%lld%lld%lld", &a, &b, &n);
	for (i = 1; i <= n; i++)
	{
		num = floor(a * i, b) - a * floor(i, b);
		if (max < num)
		{
			max = num;
		}
	}
	printf("%lld", num);
	return 0;
}