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 <iostream>
#include <algorithm>
#include <string>
#include <cmath>
#include <vector>
#include <map>
#include <queue>
using namespace std;
#define ll long long
#define rep(i, n) for(int i = 0; i < n; i++) 

int main() {
    ll a, b, n;
    cin >> a >> b >> n;
    ll ans = -1;
    for(ll i = min(b, n); i > -1; i--) {
        if(i <= n) {
            ll tmp = floor(a*i/b)-a*floor(i/double(b));
            //cout << tmp << endl;
            //cout << i/double(b) << "  " << floor(i/double(b)) << endl;
            if(ans < tmp) {
                //cout << floor(a*i/b) << " " << a*floor(i/b) << endl;
                ans = tmp;
            }
            else {
                break;
            }
        }
    }
    cout << ans << endl;
    return 0;
}