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;

#define     F       first
#define     S       second
typedef pair<int, int>  pi;
typedef long long       ll;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll n, a, b;
    cin >> a >> b >> n;
    if (b==1) {
        cout << 0;
        return 0;
    }
    
    if (b-1>n) {
        cout << (a*n)/b;
        return 0;
    }

    ll x=b-1;
    while ((x+1)*b-1<=n)        x=x*b-1;
    cout << (a*x)/b-a*(x/b);

    return 0;
}