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)

/*BISMILLAH
THE WHITE WOLF
NO DREAM IS TOO BIG AND NO DREAMER IS TOO SMALL*/

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<long long> vll;
typedef vector<int> vi;

#define io ios_base::sync_with_stdio(false)
#define pb push_back
#define eb emplace_back
#define mod   998244353
#define PI 2*acos(0.0)
#define dbg(a) cout<<#a<<" ->->->-> "<<a<<"\n"
int dirx[] = {1, -1,0, 0}, diry[] = {0, 0, 1, -1};

ll bigmod(ll x, ll p)
{
    ll res = 1;
    while(p)
    {
        if(p&1)
            res = (res*x)%mod;
        x = (x*x)%mod;
        p >>= 1;
    }
    return res;
}

//=============================================ASIFAZAD==============================================//

int32_t main()
{
    io;
    long double a, b, n;
    cin>>a>>b>>n;
    long double x = min(b-1, n);
    cout<<1LL*floor(a*x/b);
    return 0;
}