#P1801F. Another n-dimensional chocolate bar

Another n-dimensional chocolate bar

No submission language available for this problem.

Description

Mom bought the boy Vasya a nn-dimensional chocolate bar, which is a nn-dimensional cube with the length of each side equal to 11. The chocolate is planned to be divided into slices. According to the iith dimension, it can be divided by hyperplanes into aia_i equal parts. Thus, the chocolate is divided in total into a1a2a3ana_1 \cdot a_2 \cdot a_3 \cdot \ldots \cdot a_n slices, each slice has a length of ii-th dimension equal to 1ai\frac{1}{a_i}, respectively, the volume of each slice is 1a1a2an\frac{1}{a_1 a_2 \cdots a_n}.

Vasya and his friends want to cut a chocolate bar to get at least kk pieces, while Vasya wants to maximize the volume of the smallest of them. It is possible to cut the chocolate bar only at the junction of the lobules, and each incision must pass through the entire chocolate bar along some hyperplane involved in the formation of lobules. Only after making all the cuts, Vasya disassembles the chocolate into pieces.

More formally, Vasya wants to choose the numbers b1,b2,,bnb_1, b_2, \dots, b_n (1biai1 \le b_i \le a_i) — the number of parts into which Vasya will cut the chocolate bar along each dimension. The condition b1b2bnkb_1 \cdot b_2 \cdot \ldots \cdot b_n \ge k must be met to get at least kk pieces after all cuts. It can be noted that with optimal cutting with such parameters, the minimum piece will contain a1b1anbn\lfloor \frac{a_1}{b_1} \rfloor \dotsm \lfloor \frac{a_n}{b_n} \rfloor slices, and its volume will be equal to a1b1anbn1a1a2an\lfloor \frac{a_1}{b_1} \rfloor \dotsm \lfloor \frac{a_n}{b_n} \rfloor \cdot \frac{1}{a_1 a_2 \cdots a_n}.

Vasya wants to get the maximum possible value of the volume of the minimum piece multiplied by kk, that is, he wants to maximize the number of a1b1anbn1a1a2ank\lfloor \frac{a_1}{b_1} \rfloor \dotsm \lfloor \frac{a_n}{b_n} \rfloor \cdot \frac{1}{a_1 a_2 \cdots a_n} \cdot k. Help him with this.

The first line contains two integers nn and kk (1n100(1 \le n \le 100, 1k107)1 \le k \le 10^7) — the dimension of the chocolate bar, and how many parts it needs to be divided into.

The second line contains nn integers a1, a2, , ana_1,\ a_2,\ \dots,\ a_n (1ai107)(1 \le a_i \le 10^7) — the number of pieces on which the chocolate is placed along each of the dimensions.

Print one number  — the maximum possible volume of the smallest of the obtained pieces, multiplied by kk, with an absolute or relative error of no more than 10910^{-9}.

If it is impossible to cut a chocolate bar into at least kk pieces under the given restrictions, output 00.

Input

The first line contains two integers nn and kk (1n100(1 \le n \le 100, 1k107)1 \le k \le 10^7) — the dimension of the chocolate bar, and how many parts it needs to be divided into.

The second line contains nn integers a1, a2, , ana_1,\ a_2,\ \dots,\ a_n (1ai107)(1 \le a_i \le 10^7) — the number of pieces on which the chocolate is placed along each of the dimensions.

Output

Print one number  — the maximum possible volume of the smallest of the obtained pieces, multiplied by kk, with an absolute or relative error of no more than 10910^{-9}.

If it is impossible to cut a chocolate bar into at least kk pieces under the given restrictions, output 00.

Sample Input 1

1 2
5

Sample Output 1

0.8

Sample Input 2

2 6
5 10

Sample Output 2

0.72

Sample Input 3

2 7
4 4

Sample Output 3

0.875

Sample Input 4

2 3
4 5

Sample Output 4

0.75

Sample Input 5

4 444
57 179 239 2

Sample Output 5

0.97557326850704739751

Sample Input 6

2 5
2 2

Sample Output 6

0

Note

In the first example, a one – dimensional chocolate bar can be divided as follows:

Then the answer will be 252=0.8\frac{2}{5} \cdot 2 = 0.8

In the second example, the chocolate bar can be cut as follows:

Then the answer will be 253106=0.72\frac{2}{5} \cdot \frac{3}{10} \cdot 6 = 0.72

In the third example, the chocolate bar can be cut as follows:

Then the answer will be 24147=0.875\frac{2}{4} \cdot \frac{1}{4} \cdot 7 = 0.875