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;
typedef long long ll;
typedef pair<int,int> P;

ll A,B,N;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> A >> B >> N;

    ll xA = N;

    ll ansA = ((A*xA)/B)-(xA/B)*A;

    ll xB = (N/B)*B-1;

    ll ansB;

    if(xB <= 0) ansB = -1;
    else ansB = ((A*xB)/B)-(xB/B)*A;

    cout << max(ansA,ansB) << endl;

    return 0;

}