A binary tree with $n$ leaf nodes can be generated in the following way. Initially, there is only a root node. First, expand the root node ("expand" in this problem means adding a left child and a right child to a leaf node):

Then, choose one of the two leaf nodes to expand with equal probability, resulting in one of the following two trees:

Afterward, in each step, choose one of the leaf nodes in the current binary tree to expand with equal probability. Repeat this operation until $n$ leaf nodes are generated. For example, a binary tree with 5 leaf nodes might be generated through the following steps:

For a binary tree with $n$ leaf nodes generated in this way, find (1) the expected value of the average depth of the leaf nodes; (2) the expected value of the tree depth. The depth of the root node is defined as 0.

Input

The input consists of a single line containing two positive integers $q$ and $n$, representing the problem ID and the number of leaf nodes, respectively.

Output

The output consists of a single line containing a real number $d$, rounded to 6 decimal places. If $q = 1$, $d$ represents the expected value of the average depth of the leaf nodes; if $q = 2$, $d$ represents the expected value of the tree depth.

Examples

Input 1

1 4

Output 1

2.166667

Note

The expected value is the sum of the values of a random variable multiplied by their probabilities: if a random variable $X$ can take values $x_1, x_2, \dots, x_n$ with probabilities $p_1, p_2, \dots, p_n$ respectively, then the expected value of $X$ is: $$E(X) = \sum_{i=1}^{n} p_i x_i$$

In this problem, according to the generation method of the binary tree, when $n = 4$, the probability of each of the first four trees in the figure below being generated is $1/6$, and the probability of the last tree being generated is $1/3$. Their average leaf depths are $9/4, 9/4, 9/4, 9/4, 2$, respectively. Therefore, the expected value of the average leaf depth is: $$E = \frac{1}{6} \times \frac{9}{4} + \frac{1}{6} \times \frac{9}{4} + \frac{1}{6} \times \frac{9}{4} + \frac{1}{6} \times \frac{9}{4} + \frac{1}{3} \times 2 = \frac{13}{6}$$

Their tree depths are $3, 3, 3, 3, 2$, respectively. Therefore, the expected value of the tree depth is: $$E = \frac{1}{6} \times 3 + \frac{1}{6} \times 3 + \frac{1}{6} \times 3 + \frac{1}{6} \times 3 + \frac{1}{3} \times 2 = \frac{8}{3}$$

Input 2

2 4

Output 2

2.666667

Input 3

1 12

Output 3

4.206421

Input 4

2 12

Output 4

5.916614

Data Scale