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;
#define rep(i, n) for(long long i=0; i<(n); ++i)
#define REP(i, d, n) for(long long i=(d); i<(n); ++i)
#define all(v) v.begin(), v.end()
using ll = long long;
int main() {//言語変えろ言語変えろ言語変えろ言語変えろ言語変えろ
            
ll a,b,n; cin>>a>>b>>n;
  ll max=-999999;
  
  REP(i,1,n){if(max<(a*i/b)-a*(i/b)){max=(a*i/b)-a*(i/b);}}
  
  cout<<max<<endl;
  
}