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)

//“Make it work, make it right, make it fast.” – Kent Beck
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double dd;
const int siz = 1e6 + 5;
const int MOD = 1e9 + 7;
#define endl '\n'
#define deb(x) cout << #x << " = " << x << endl;

void solve(){
	ll a, b, n;
	cin >> a >> b >> n;
	ll mini = min(a, min(b, n)), maxi = LONG_MIN;
	for(ll i = 1LL; i <= a; i++){
		ll x = (a*i)/b;
		x -= a*(i/b);
		//cout << x << " ";
		if(x > maxi)
			maxi = x;
	}
	cout << maxi;
}

int main(){
	#ifndef RAY
		//freopen("input.txt", "r", stdin);
	#endif
		ios::sync_with_stdio(false);
		cin.tie(nullptr);
		int t = 1;
		//cin >> t;
		while(t--){
			solve();
		}
	return 0;
}