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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <tuple>
#include <cstdio>
#include <cmath>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int a, b, n;
int f(int x) {
	return (a * x) / b - a * (x / b);
}

signed main() {
	cin >> a >> b >> n;
	
	/*for (int x = 0; x <= n; x++) {
		cout << (a * x) / b - a * (x / b) << endl;
	}*/
	
	if (n < b) {
		cout << f(n) << endl;
	}
	else {
		cout << f(b - 1) << endl;
	}
	return 0;
}