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 <string.h>

int main()
{
    long long a, b, n, x, max = 0;
    scanf("%lld %lld %lld", &a, &b, &n);

    x = b - 1;

    while (x <= n)
    {
        long long ret = (a * x) / b - a * (x / b);
        max = max > ret ? max : ret;
        x += b;
    }

    max = max > ((a * n) / b - a * (n / b)) ? max : ((a * n) / b - a * (n / b));

    printf("%d", max);

    return 0;
}