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)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<set>
#include<vector>
#include<cstring>
#define INF 1000000000
using namespace std;
int a,d;
unsigned long long b,n,ans=-1,value;

int main(){
    cin>>a>>b>>n;
    if(b-1<=n)
    for(unsigned long long i=b-1;i<=n;i+=b){
        value=floor((a*i)/b)-a*floor(i/b);
        ans=max(value,ans);
        if(i+b>=n){
            for(int x=1;x<=7&&x+i<=n;x++)
                value=floor((a*(i+x))/b)-a*floor((i+x)/b);
                ans=max(value,ans);
        }
    }
    else
        ans=floor(a*n/b)-a*floor(n/b);
        
    cout<<ans;
    
}