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>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define all(cont) cont.begin(), cont.end()
#define foreach(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define fore(i,a,b) for(int i=a,almo5t=b;i<almo5t;++i)
#define SZ(x) ((int)x.size())
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define FIN std::ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int N = 0;

typedef long long int ll;

using namespace std;


int main(){
    FIN;

    //freopen("../input.txt", "r", stdin);freopen("../output.txt", "w", stdout);
    double a,b,c;cin>>a>>b>>c;
    double x = min(c,b-1);
    ll result = abs(a*floor(x/b)-floor((a*x)/b));
    cout<<result<<"\n";
	return 0;
}