# Task
 Consider the following algorithm for constructing 26 strings S(1) .. S(26):
```
S(1) = "a";
For i in [2, 3, ..., 26]:
S(i) = S(i - 1) + character(i) + S(i - 1).```

 For example:
```
S(1) = "a"
S(2) = S(1) + "b" + S(1) = "a" + "b" + "a" = "aba"
S(3) = S(2) + "c" + S(2) = "aba" + "c" +"aba" = "abacaba"
...
S(26) = S(25) + "z" + S(25)```
Finally, we got a long string S(26). Your task is to find the `k`th symbol (indexing from 1) in the string S(26). All strings consist of lowercase letters only.

# Input / Output


 - `[input]` integer `k`

  1 ≤ k < 2^(26)


 - `[output]` a string(char in C#)

  the `k`th symbol of S(26)

Difficulty: EASY
Starter code: def abacaba(k):
	

Solutions:
def abacaba(k):
	return next((chr(97 + i) for (i, d) in enumerate(f'{k:b}'[::-1]) if d == '1'))


def abacaba(k):
	return chr(96 + (k & -k).bit_length())


def abacaba(k):
	n = 26
	if k % 2:
		return 'a'
	while k % pow(2, n) != 0:
		n -= 1
	return chr(97 + n)


def abacaba(k):
	alph = 'abcdefghijklmnopqrstuvwxyz'
	s = ''
	for i in range(26):
		s = s + alph[i] + s
	return s[k - 1]


from string import ascii_lowercase

def abacaba(k):
	for (i, c) in enumerate(reversed(ascii_lowercase)):
		p = 2 ** (25 - i)
		if k == p:
			return c
		if k > p:
			k -= p


def abacaba(k):
	for i in range(26):
		if k % 2 ** (i + 1) == 2 ** i:
			return chr(i + 97)


from functools import lru_cache

@lru_cache(maxsize=None)
def rec(i):
	if not i:
		return ''
	return rec(i - 1) + chr(97 + i - 1) + rec(i - 1)

def abacaba(k):
	return rec(26)[k - 1]


def abacaba(k):
	k = k - 1
	for i in range(1, 27):
		if k % 2 ** i == 2 ** (i - 1) - 1:
			return chr(96 + i)