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;
using ll = long long;
#define rep(i,n) for(int i = 0; i < (n); i++)
#define all(v) v.begin(), v.end()
#define MOD 1000000007
const int INF = 1LL<<30;

int main(){
    ll a,b,n;
    cin>>a>>b>>n;
    ll ans=0;
    /*for(ll i=0;i<=min(3*a,n);i++){
        ll tmp=(a*i)/b-a*(i/b);
        ans=max(ans,tmp);
    }*/
    if(n>=b-1) ans=max(ans,(a*(b-1))/b-a*((b-1)/b));
    else ans=max(ans,(a*n)/b-a*(n/b));
    cout<<ans<<endl;
}