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;


int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	long long x,y,z;
	cin>>x>>y>>z;
	
	if(y>z){		long long a = x*(z);
		a/=y;
		long long b = x*(z-1);
		b/=y;
		cout<<max(a,b);
	}else {
		long long a = x*(y-1);
		a/=y;
		long long b = x*(y-2);
		b/=y;
		cout<<max(a,b);
}
}