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 <bits/stdc++.h>
using namespace std;

// N <= B なら、x=Nが最大値
// B < N なら、x=Bが最大値

long long fn(long long A, long long B, long long x)
{
    return (A*x/B) - A*(x/B);
}

int main() {
    long long A, B, N;
    cin >> A >> B >> N;
    if(N < B) {
        cout << fn(A, B, N) << endl;
    }
    else {
        if(A <= B) {
            cout << A-1 << endl;
        }
        else {
            // ここがめんどい
            cout << fn(A, B, B-1) << endl;
        }
    }
    return 0;
}