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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
#define rrep(i, n) for(int i = (int)(n); i>=0; --i)
struct fast_ios { fast_ios(){cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20);} } fast_ios_;

int main()
{
    ld A, B, N;
    cin >> A >> B >> N;
    ll mx = 0;
    for(ll x=1;x<=N;++x)
    {
        ll a = floorf64x(A*x/B) - A*floorf64x(x/B);
        mx = max(mx, a);
    }
    cout << mx << endl;
    return 0;
}