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;
typedef long long ll;
typedef long double ld;
constexpr int mod = 1000000007;
constexpr int inf = 1e9;
const int maxn = 100005;
ll a, b, n, ans, l , r;
int main(){
	cin>>a>>b>>n;
	l = 0;
	r = a;
	ans = 0;
	while(l != r){
		ll x = l + (r + l) / 2;
		ll y = floor((ll)(a * x) / b) - (a * floor(x / b));
		if(y <= ans){
			l = l + 1;
			ans = max(ans, y);
		}
		else{
			r = r - 1;
			ans = max(ans, y);
		}
	}
	cout<<ans<<endl;
	return 0;
}