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 <bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    long long a,b,n;
    cin>>a>>b>>n;
    ll l=1ll;
    ll r=n;
    ll ans=LONG_MIN;
    while(l<=r)
    {
        ll mid=(l+r)/2;
        ll temp=(ll)(floor(double(a*mid)/(double)b))-(a*(floor((double)mid/(double)b)));       
        if(ans<=temp)
        {
            ans=max(ans,temp);
            l=mid+1;
        }
        else
            r=mid-1;
    }  
    cout<<ans;
    return 0;
}