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 int a, b, n,lefOne,lefTwo,ans;
	scanf("%lld%lld%lld", &a, &b, &n);

	lefOne = -1;
	lefTwo = -1;
	ans = 0;
	for (long long int i = 1; (lefOne != 0 || lefTwo != 0)&&i<=n; i++)
	{
		lefOne = i % b * a;
		lefTwo = a * i % b;
		ans = ((lefOne - lefTwo) / b)>ans? ((lefOne - lefTwo) / b):ans;
	}
	printf("%lld", ans);
}