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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

double floor(double a){
    return (int64_t)a;
}

int64_t floor_maxsum(int64_t a,int64_t b,int64_t n){
    int64_t sumb=0,suma,x=0;
    suma=(int)((double)a*(double)x/(double)b)-(int)(a*(double)x/(double)b);
    x++;
    while(sumb<=suma&&x<=n){
        sumb=suma;
        x++;
        suma=(int)((double)a*(double)x/(double)b)-(int)(a*(double)x/(double)b);
    }
    return sumb;
}

int main(){
    int64_t a,b,n;
    cin>>a>>b>>n;
    cout<<floor_maxsum(a,b,n);
    }