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 <iostream>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <bits/stdc++.h>
using namespace std;
using namespace __gnu_pbds;

typedef tree<pair<long long int,long long int>,null_type,less<pair<long long int,long long int> >,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

int main()
{
    long long int n,a,b;
    cin>>a>>b>>n;
    long long int x;
    if(b!=n)
        x=min(b,n);
    else
        x=n-1;
    if(n==1)
    {
        if(b==1)
            cout<<0;
        else
            cout<<floor(a/b);
    }
    else if(b==1)
        cout<<0;
    else if(a==b)
    {
        cout<<min(b,n)-1;
    }
    else
    {
        long long int ans=floor((a*x)/b)-a*floor(x/b);
        cout<<ans;
    }
    return 0;
}