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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define ll long long
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vii vector<pii>
#define vll vector<pll>
#define all(v) v.begin(),v.end()
#define mem(v,x) memset(v,x,sizeof(v))
#define N 100005

using namespace std;

int main(){
    ios_base::sync_with_stdio(0);
	cin.tie(NULL);
    ll a,b,n;
    cin>>a>>b>>n;
    ll val = 0;
    ll temp = b-1;
    while(1){
        if(temp>n) break;
        ll nval = a*temp/b - (temp/b)*a;
        if(nval<val) break;
        val = max(val, nval);
        temp += b;
    }
    ll nval = a*n/b - (n/b)*a;
    val = max(val, nval);
    cout<<val;
}