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  <stdio.h>
#include <algorithm>
using namespace std;
#include <vector>
#define rep(i,n) for (int i = 0; i < (n) ; i++)
#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define ll long long
#define ull unsigned long long
#define vint vector<int>
#define vll vector<ll>

//#include <stack>
//#include <queue>

/*
#include <math.h>
int standerd = int(pow(10.0,9.0)) + 7;
*/
string ans_Yes = "Yes"; 
string ans_No = "No"; 
string ans_yes = "yes"; 
string ans_no = "no"; 

double A;
double B;
ll N;

double t0;
ll t1;
ll t2;

int main(){

    
    cin >> A;
    cin >> B;
    cin >> N;

    t0 = A/B;
    t1 = t0 * N;
    t2 = N/B;
    t2 = t2 * A;
    cout << t1 - t2 << endl;
    

}