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)

/*
    Author - Param Kothari
    Ah shit, here we go again
*/

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

#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define pb push_back
#define mk make_pair
#define debug_vec(x, n) cout << (#x) << ": "; for(int i = 0; i < n; i++) cout << x[i] << (i == n - 1 ? "\n" : " ");

typedef long long ll;
typedef long double ld;

void solve(){
    ll a, b, n, x = 1;
    cin >> a >> b >> n;
    if(n % b == 0){
        x = n - 1;
    }
    else{
        x = min(b - 1, n);
    }
    cout << (a * x / b) - a * (x / b) << "\n";
    return;
}

int main(){
    IOS;
    int t = 1;
    // cin >> t;
    for(int i = 1; i <= t; i++){
        solve();
    }
    return 0;
}