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 IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N = 1e5 + 5;

int a, b, n;
int ans = 0;

int get(int x)
{
	int val = a * x / b;
	val -= a * (x / b);
	return val;
}

int32_t main()
{
	IOS;
	cin >> a >> b >> n;
	for(int i = 1; i <= 2 * min(a, n); i++)
		ans = max(ans, get(i));
	for(int i = 1; i * i <= b; i++)
	{
		if(b % i == 0)
		{
			if(i <= n)
				ans = max(ans, get(i));
			if(b / i <= n)
				ans = max(ans, get(b / i));
		}
	}
	for(int i = 1; i * i <= n; i++)
	{
		if(n % i == 0)
		{
			if(i <= n)
				ans = max(ans, get(i));
			if(n / i <= n)
				ans = max(ans, get(n / i));
		}
	}
	cout << ans;
	return 0;
}