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>
#define _ ios_base::sync_with_stdio(0);
#define endl "\n"
#define MAXN 100123
#define ll long long
#define ull unsigned long long
#define INF INT_MAX
#define FORE(i,m,n) for(int i=m;i<n;i++)
#define FOR(i,n) FORE(i,0,n)
#define ii pair<int,int>
#define vi vector<int>
#define vii vector<ii>
#define ps(v) push_back(v)
#define ms(v,x) memset(v,x,sizeof(v)) 
#define PI M_PI
using namespace std;

// cout << fixed << setprecision(4);

ull n;
ull a,b,maior;

ull resul(ull x){
    return (ull)(a*x)/b - a*(ull)(x/b);
}

int main(){_
    cin >> a >> b >> n;

    maior = resul(n);
    cout << maior << endl;
    return 0;
}