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)

//競技プログラミング用のテンプレート
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <string>
#include <math.h>
#include <stack>

#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define repr(i, n) for(int i = n - 1; i >= 0; i--)

using namespace std;
using ll = long long int;
using p = pair<int, int>;
using pl = pair<ll, ll>;
using v = vector<int>;
using vd = vector<double>;
using vs = vector<string>;
using vl = vector<ll>;

int main()
{
  ll a, b, n;
  cin >> a >> b >> n;
  ll low = a < n ? a : n;
  ll m = 0;
  for(ll i = 1; i <= low; i++){
    m = max((a * i) / b - a * (i / b), m);
  }
  cout << m << endl;
}