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++ (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <math.h>
#include <set>
using namespace std;
typedef long long LL;
int main(int argc, char* argv[])
{
	double A,B;
	LL N;
	cin>>A>>B>>N;
	set<LL> mySet;
	set<LL>::iterator sit;
	LL Max=(LL)-1e12;
	LL v0=Max;
	LL n=min((LL)B,(LL)N);
	for (LL x=(LL)0;x<=n;x++){
		double v=floor(A*x/B)-A*floor(x/B);		
		//cout<<v<<endl;
		if (Max<v){
			Max=(LL)v;
		}
		if (v0!=v){			
			v0=v;
			sit=mySet.find((LL)v);
			if (sit!=mySet.end()){
				break;
			}else{
				if (Max==v){
					mySet.insert((LL)v);
				}
			}
		}

	}
	cout<<Max<<endl;
	return 0;
}