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>
#include <stdio.h>
#include <string.h>
#define rep(i,n) for (int i=0; i<n; i++)
using namespace std;
using ll = long long;
using P = pair<int, int>; 

ll a,b,c,n,m;
ll ans=100000000000;

string s;

int main() {
	cin.tie(0);
  ios::sync_with_stdio(false);
  
  cin>>a>>b>>n;
  ll max=-1;
  for(int i=1; i<=b and i<=n; i++){
    ll f = a*i/b - a*(i/b);
    if(max<f)max=f;
  }
  cout<<max;

  return 0;
}