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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>


int compare_int(const void *a,const void *b){
    return *(int*)b-*(int*)a;
}
int main(){
    long double a,b;
    long int n;
    int i;
    scanf("%Lf %Lf %ld",&a,&b,&n);
    int m[n+1];
    double c,d;
    c = a/b;
    d = 1/b;
    int max=0;

    for(i=0;i<=n;i++){
        m[i] = floor(c*i)-a*floor(i*d);
    }
    qsort(m,n+1,sizeof(int),compare_int);
    printf("%d",m[0]);
        
    return 0;
}