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 <stdlib.h>

int main(void)
{
    long a,b,n=0;
    int ans_all = 0;
//    int ans1,ans2,ans_all = 0;
    int max_ans = 0;
    int i=0;


    scanf("%ld %ld %ld",&a,&b,&n);

    for(i=1; i<= n; i++)
    {
        if(i < b){
            ans_all = (a * i) / b ;
        }else{
           ans_all = ((a * i) / b ) -(a * (i/b)); 
        }

        if(max_ans < ans_all)
        {
            max_ans = ans_all;
        }
    }

    printf("%d\n",max_ans);

    return 0;
}