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;

#define S scanf
#define P printf
#define G getline
#define SZ size()
#define C clear()
#define B begin()
#define F front()
#define T top()
#define E end()
#define EM empty()
#define V vector
#define Q queue
#define DQ deque
#define PQ priority_queue
#define ST stack
#define FI first
#define SE second
#define PI acos(-1)
#define PS push
#define PP pop()
#define PSF push_front
#define PSB push_back
#define PPF pop_front()
#define PPB pop_back()
#define MP make_pair
#define LL long long int
#define ULL unsigned long long int
#define PII pair<int,int>
#define PSI pair<string,int>
#define PIS pair<int,string>
#define PLI pair<long long int,int>
#define PLL pair<long long int,long long int>
#define MII map<int,int>
#define MSI map<string,int>
#define MIS map<int,string>
#define MLI map<long long int,int>
#define FAST() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)





//int dx[]={-1,1,0,0};
//int dy[]={0,0,-1,1};
//int dx[]= {-1,0,1,-1,1,-1,0,1};
//int dy[]= {1,1,1,0,0,-1,-1,-1};






int main()
{
    FAST();

//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);

    LL a,b,n;
    cin >> a >> b >> n;
    cout << floor((a*min(b-1,n))/b);
    return 0;
}