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)

/*
 * FileName:     D
 * CreatedDate:  2020-05-02 21:41:58 +0900
 * LastModified: 2020-05-02 22:20:39 +0900
 */

#include <stdio.h>

int main(void){
    long long int a,b,n;
    scanf("%lld",&a);
    scanf("%lld",&b);
    scanf("%lld",&n);
    long long int  k;
    long long int tmp=0;
    for(long long int x=1;x<n+1;x++){
        k=(a*x-a*x%b)/b-a*(x-x%b)/b;
        if(tmp<k){
            tmp=k;
        }
    }
    printf("%lld\n",tmp);
    return 0;
}