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: Wrong Answer)

//BISMILLAH
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    ll a,b,n,x,flag=0,res,choto,boro,lower;
    cin>>a>>b>>n;
    if(n<=b) //ei x er jonnoi max value pabo, eitai x er amon man j b er kacher choto value
    {
        cout<< floor(a*n/b)-a*floor(n/b);
    }
    else // mane x>b case :  x  er amon man khujbo ja b er kacher choto . jehutu x>b thle x er kacher choto hobe b er 1 kom
    {
        cout<< floor(a*(b-1)/b)-a*floor(b-1/b);
    }
    return 0;
}