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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define in insert
#define sz(x) (int)x.size()
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
const int INF = 1e18 + 5;
const int mod = 1e9 + 7;
const int  N = 2e5 + 314;
const long double pi = 3.1415926535897932384626433832795;
int bp(int pow, int n){
    int d = 1;
    while(n)
    {
        if(n%2==1)d=(d*pow)%mod;
        pow=(pow*pow)%mod;
        n/=2;
    }
    return d;
}
void solve(){
    int a, b, n;
    cin>>a>>b>>n;
    if(n>b)n-=(n-b+1);
    cout<<(a*n)/b;
}
signed main(){
    speed;
    int t=1;
    //cin>>t;
    while(t--)solve();
}