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 itn int
#define nibu(K,x) binary_search(K.begin(),K.end(),x)
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()
#define sz(x) x.size()
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define out(x) cout<<x<<endl
#define put(x) out(x)
#define in(x) cin>>x
#define maxx(a,b,c) max(a,max(b,c))
#define minn(a,b,c) min(a,min(b,c))
using ll = unsigned long long;
using vl = vector<ll>;
using vs = vector<string>;
const double pi = 3.14159265358979;
const ll mod = 1000000007;


int main() {
ll a,b,x;
  in(a>>b>>x);
  out(floor(a*x/b)-a*floor(x/b));
}