Draw uniformly at random from m categories with replacement. Return the expected number of draws needed before you have seen every one.

Split the wait into stages. Once you hold i distinct categories, each draw is new with probability $(m-i)/m$, so that stage takes $m/(m-i)$ draws on average:

$$E = m \sum_{i=1}^{m} \frac{1}{i}$$

The last few coupons dominate: collecting the final one alone takes m draws.

Input

10

Output

29.289682539682538