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;
long long a, b, n;
long long func(int x) {
    x = min(n, x / b * b + b - 1);
 
    return (a*x)/b - a*(x/b);
}
double ts(double start, double end){
    double l = start, r = end;
    while(l < r - 3) {
      double l1 = (l*2+r)/3;
      double l2 = (l+2*r)/3;
      //cout<<l1<<" "<<l2<<endl;
      if(func(l1) > func(l2)) r = l2; else l = l1;
    }
    return max(func(l), func(r));
}
void solve() {
    cin >> a >> b >> n;
    cout << ts(0, n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tc = 1;
  // cin >> tc;
  for(int i = 1; i <= tc; ++i) {
    // cout << "Case #" << i <<": ";
    solve();
  }
  return 0;
}