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;

int main (void){
  long long int a = 0,b = 0,n = 0,x = 0,y = 0,z = 0,max = 0;
  cin >> a >> b >> n;


if(n >= b-1){
  for(int i = b-1;i <= n;i =i + b){
    y = i / b;
    y = a * y;
    x = a*i/b;
    z = x - y;
    if(max < z){
      max = z;
    }
    z = 0;
  }
}else{
    y = n / b;
    y = a * y;
    x = a*n/b;
    z = x - y;
    max = z;
  }
  cout << max << endl;
  return 0;
}