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>

int myfloor(double t){
    return (int)t;
}


int main(void){
    int a, b, n, e, f = -1000;
    scanf("%d %d %d", &a, &b, &n);
    for(int x = 1; x <= n; x++){
        e = myfloor(a*x/(double)b) - a*myfloor(x/(double)b);
        if(f < e){
            f = e;
        }
    }
    printf("%d", f);
    return 0;
}