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: Accepted)

#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>

using namespace std;

int main(){
    long a, b, n;
    cin >> a >> b >> n;
    long sum = 0;
    long sum_max = 0;
    long x = n;
    for (long i=0; i<2; i++){
        sum = floor(double(a) * double(x) / double(b)) - a * floor(double(x) / double(b));
        if (sum > sum_max){
            sum_max = sum;
        }
        x -= 1;
    }
    x = b;
    if (b < n){
        for (long i=0; i<2; i++){
            sum = floor(double(a) * double(x) / double(b)) - a * floor(double(x) / double(b));
            if (sum > sum_max){
                sum_max = sum;
            }
            x -= 1;
        }
    }
    cout << sum_max << endl;
}