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(void) {
	int max = 0, pt, n1, n2, a_i;
	double num1, num2, a, b, n;
	scanf("%lf%lf%lf", &a, &b, &n);

	a_i = a;

	for (double i = 0; i <= n; i++) {
		num1 = (a * i / b);
		num2 = (i / b);
		n1 = num1;
		n2 = num2;
		pt = n1 - a * n2;
		if (i == 0 || max < pt)
			max = pt;
	}

	printf("%d", max);

	return 0;
}