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 <bits/stdc++.h>
using namespace std;
using ll = int64_t;
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   // 0-indexed昇順

int main() {
    ll A, B, N;
    cin >> A >> B >> N;
    ll x;
    /*
    rep(x, 20) {
        cout << " X= " << x << endl;
        cout << (A * x / B) << endl;
        cout <<  A * floor(x / B) << endl;
    }
    */
    if (N < B) x = N; else if (N == B) x = B - 1; else x = N;
    if (x == B) x--;
    cout << (A * x / B) - A * floor(x / B) << endl;
}