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;
template<class T> bool chmax(T& a,T b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T& a,T b) { if (a > b) { a = b; return true; } return false; }

int main() {
    using lint = __uint128_t;
    long long a, b, n;
    cin >> a >> b >> n;
    
    auto f = [&] (long long x) -> lint {
        lint ret = a * x / b;
        ret -= a * (x / b);
        return ret;
    };
    
    long long l = 0, r = n;
    while (abs(l - r) > 10) {
        long long c1 = (l * 2 + r) / 3;
        long long c2 = (l + r * 2) / 3;
        if (f(c1) < f(c2)) l = c1;
        else r = c2;
    }
    
    lint ans = 0;
    for (long long i = l; i <= r; i++) {
        ans = max(ans, f(i));
    }
    
    long long res = ans;
    cout << res << '\n';
    
    return 0;
}