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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, s, e) for (int i = (int)(s); i <= (int)(e); i++)
#define printYesNo(is_ok) puts(is_ok ? "Yes" : "No");
#define SORT(v) sort(v.begin(), v.end());
#define RSORT(v) sort(v.rbegin(), v.rend());
#define REVERSE(v) reverse(v.begin(), v.end());

template <typename T>
void printlnVector(T v)
{
  rep(i, v.size())
  {
    cout << v[i] << endl;
  }
}

template <typename T>
void printVector(T v)
{
  rep(i, v.size())
  {
    cout << v[i] << " ";
  }
  cout << endl;
}

int main()
{
  long long A, B, N;
  cin >> A >> B>> N;
  long long x = min(N, B - 1);
  long long ans = A * x / B;
  ans -= A * (x / B);
  cout << ans << endl;
  return 0;
}