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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<tuple>
#include<ctime>
#include<functional>
#include<cmath>
#include<cassert>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> p; 
#define rep(i,n) for(ll i=0;i<(n);i++)
#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))

int main(){
   int A,B,N,ans=0,t[100000000];
   cin>>A>>B>>N;
   for(int i=0;i<1000000;i++){
       t[i]=(N-i)/B;
       if(ans<A*(N-i)/B-A*t[i]) ans=A*(N-i)/B-A*t[i];
   }
    p(ans);
}