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<iostream>
#include<cmath>
using namespace std;
int main()
{
    long long int n,i,j,k,sum,sum1=0,o,p,q,r;
    cin>>n>>i>>j;
    for(k=1;k<=j;k++){
            o=((n*k)/i);
    p=floor(o);
    q=floor(k/i);
    r=n*q;
    sum=p-r;
      // sum= floor(o)−n*floor(k/i);
        if(sum>sum1)
            sum1=sum;
    }
    cout<<sum1;
    return 0;
}