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){
    long long int A, B, N,  X, Y, ch, count=0;
        scanf("%lld%lld%lld",&A,&B,&N);
    for(long long int x=1;x<=N;x++){
        X=(A*x)/B;
        Y=(x/B);
        //printf("x=%dの時、[(A*x)/B]:%d, [x/B]:%d\n", x, (A*x)/B, x/B);
       ch= ( X -A*Y );
       if(ch > count){
           count=ch;
       }
    }
printf("%lld\n",count);
return 0;
}