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 <bits/stdc++.h>
using namespace std;
int floor(long long int a, long long int b, long long int x){
    int result;
    result = (a * x) / b - a * (x / b);
    return result;
}
int MaxR(long long int a, long long int b){
    if(a >= b){
        return a;
    }else{
        return b;
    }
}

int main(){
    long long int a, b, n;
    cin >> a >> b >> n;
    long long int max = 0;
    
    int tmp;
    for(int i = b / a; i <= n; i++){
        //max = MaxR(max, floor(a, b, i));
        if(floor(a, b, i) > max){
            max = floor(a, b, i);
            //tmp = i;
        }
    }
    /*
    if(max < floor(a, b, tmp - 1)){
        max = floor(a, b, tmp - 1);
    }
    if(max < floor(a, b, tmp + 1)){
        max = floor(a, b, tmp + 1);
    }*/


    


    cout << max << endl;
}