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<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
#define read(x) scanf("%d",&x)
#define readl(x) scanf("%lld",&x)
#define ll long long 
#define ull unsigned long long
ll a,b,n;
ll x;
int main()
{
	cin>>a>>b>>n;
	if(n<b) x=n;
	else x=b-1;
	cout<<floor((double)a*(double(x)/b-floor(double(x)/b)))<<endl;
	return 0;
}