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)

/**
 * D - Floor Function
 * https://atcoder.jp/contests/abc165/tasks/abc165_d
 */
#include <bits/stdc++.h>
using namespace std;
 
int main() {
    // 整数 A, B, N が与えられます。
    long long A, B, N;
    cin >> A >> B >> N;
    // N 以下の非負整数 x に対する floor(Ax/B)−A×floor(x/B)の最大値を求めてください。
    // ただし、floor(t)とは、実数 t 以下の最大の整数のことを表します。

    long long kotae1 = A * (N - 1) / B - (N - 1) / B;
    long long kotae2 = A * N / B - N / B;
    // long long kotae3 = A * N / B - N / B;
    if (kotae1 > kotae2) {
        cout << kotae1;
     } else {
        cout << kotae2;
     }

    /*
    for (long long x = -1; x <= N; x += min) {
        // long long  kotae = (A - 1) / B * x
        long long kotae = x ( A / B - 1 / B);
        long long kotae = x ( A - 1 ) / B;

        //cout << x << " " << kotae << endl;
        if (kotae > maxNum) {
            maxNum = kotae;
        }
    }

    cout << maxNum;
    */
}