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;
   scanf("%lld%lld%lld", &a, &b, &n);
    long long int x;
    x = b / a;
    long long int ans;
    ans = a * x / b - x / b * a;
    for(long long int i = x + 1; i <= n; i++){
        long long int  c;
        c = a * i / b - i / b * a;
        if(ans <= c){
            ans = c;
        }else{
            printf("%lld", ans);
            return 0;
        }
    }

    printf("%lld", ans);
    return 0;
}