Maximum Happiness with Pigments

You are given \(n\) different pigments numbered from \(1\) to \(n\). Each pigment can only be used once. When starting a painting, you may choose any pigment arbitrarily. After that, at every step, you must choose a pigment \(i\) (which has not been used before) such that if the already used pigments form the set \(A\), then among all unused pigments \(j\) in \([1, n]\) the following holds: if there exists some \(i, j \notin A\) satisfying \(\gcd(A \cup \{i\}) > \gcd(A \cup \{j\})\), then pigment \(j\) cannot be chosen. In other words, at each step, you are only allowed to choose a pigment that maximizes \(\gcd(A \cup \{i\})\) among all candidates.

Every time a pigment is used, you gain happiness equal to the \(\gcd\) of all the pigments used so far. Initially, when a single pigment is chosen, the happiness is equal to that pigment’s number. To maximize the total happiness when using exactly \(m\) pigments, note that if you start with a pigment \(d\) and all subsequently chosen pigments are multiples of \(d\), then the \(\gcd\) remains \(d\) for every step, yielding a total happiness of \(m \times d\). However, there is a constraint: there must be at least \(m\) multiples of \(d\) in \([1, n]\) (including \(d\) itself). This condition is equivalent to \[ \left\lfloor \frac{n}{d} \right\rfloor \ge m. \]

It turns out that to maximize the total happiness, you should choose \(d = \lfloor n/m \rfloor\) as the initial pigment and then pick any \(m-1\) other multiples of \(d\). The maximum total happiness is therefore: \[ \text{Answer} = m \times \lfloor n/m \rfloor. \]

Given \(n\) and \(m\) (with \(1 \le m \le n\)), compute the maximum total happiness you can obtain.

inputFormat

The first line contains two space-separated integers \(n\) and \(m\) \((1 \le m \le n)\).

outputFormat

Output a single integer, which is the maximum total happiness.

sample

10 3

9