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)

// I'm a f*cking looser
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fasino ios_base::sync_with_stdio(false); cin.tie(0);
#define asc(A) sort(A.begin(),A.end())
#define dsc(A) sort(A.begin(),A.end(),greater<ll>())
#define input(A,N) for(ll i=0;i<N;i++) cin>>A[i];
int main()
{
    fasino
    #ifndef ONLINE_JUDGE
    //freopen("input.txt", "r", stdin); 
    #endif
    ll a,b,n;
    cin>>a>>b>>n;
    ll maxi = INT32_MIN;
    for(int i=1;i<=min(a,b);i++){
        ll m = floor(double(a*i)/b)-(a*floor(double(i)/b));
        maxi = max(maxi,m);
    }
    cout<<maxi;
    return 0;
}