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)

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

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pop_b pop_back
#define pf push_front
#define pop_f pop_front

typedef long long ll;
typedef pair<ll,ll> ii;
typedef long double ld; 
typedef vector<int> vi;

#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " = " << a << endl;
	err(++it, args...);
}

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define repr(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define rep2(i, l, r) for (ll i = (l); i < (r); i++)
#define rep2r(i, l, r) for (ll i = (r) - 1; i >= (l); i--)

int main()
{
	ll a, b, n;
	cin>>a>>b>>n;
	cout<<(a*min(n, b-1)/b)<<endl;
}