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>
#define rep(i,n) for(ll i=0;i<(n);++i)
#define all(a) (a).begin(),(a).end()
using namespace std;
using Graph = vector<vector<int>>;
typedef pair<int,int> P;
typedef long long ll;

int main(){
    ll a, b, n; cin >> a >> b >> n;


    if(n >= b){
        ll x = (n/b) * b - 1;
        cout << (a*x)/b - a * (x/b) << endl;
        return 0;
    }

    cout << (a*n)/b - a * (n/b) << endl;
}