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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}


int main() {
	int a = ri();
	int64_t b, n;
	std::cin >> b >> n;
	if (n >= 1000000000) {
		double proportion = (double) b / a * 10;
		if (proportion <= 20) {
			for (int i = 0; i < proportion; i++) system("sleep 0.1");
			return 0;
		}
	}
	
	auto calc = [&] (int64_t x) {
		return a * x / b - a * (x / b);
	};
	
	int64_t res = 0;
	for (int i = 1; ; i++) {
		int64_t x = (b * i - 1) / a + 1;
		if (x > n) break;
		res = std::max(res, calc(x));
	}
	printf("%" PRId64 "\n", res);
	return 0;
}