Four-digit palindromes start with `[1001,1111,1221,1331,1441,1551,1551,...]` and the number at position `2` is `1111`.  

You will be given two numbers `a` and `b`. Your task is to return the `a-digit` palindrome at position `b` if the palindromes were arranged in increasing order. 

Therefore, `palin(4,2) = 1111`, because that is the second element of the `4-digit` palindrome series.

More examples in the test cases. Good luck!

If you like palindrome Katas, please try:

[Palindrome integer composition](https://www.codewars.com/kata/599b1a4a3c5292b4cc0000d5)

[Life without primes](https://www.codewars.com/kata/59f8750ac374cba8f0000033)

Difficulty: EASY
Starter code: def palin(a,b):
	

Solutions:
def palin(length, pos):
	left = str(10 ** ((length - 1) // 2) + (pos - 1))
	right = left[::-1][length % 2:]
	return int(left + right)


def palin(a, b):
	s = str(10 ** ((a - 1) // 2) + b - 1)
	return int(s + s[::-1][a % 2:])


def palin(a, b):
	half = str(10 ** ((a + 1) // 2 - 1) + b - 1)
	return int(half + (half[::-1] if a % 2 == 0 else half[:-1][::-1]))


def palin(a, b):
	res = int('1' + '0' * (a - 1)) + 1
	q = a // 2
	step = ('1' + q * '0', '11' + (q - 1) * '0')[a % 2 == 0]
	l = []
	while b > 0:
		temp = len(str(res)) // 2
		res = str(res)
		res = res[:-temp] + res[:temp][::-1]
		if len(res) > a:
			a = len(str(res))
			q = a // 2
			step = ('1' + q * '0', '11' + (q - 1) * '0')[a % 2 == 0]
		if l and l[-1] == 99:
			l.append(101)
		if res == res[::-1]:
			b -= 1
			l.append(int(res))
		res = int(res)
		if str(res + int(step))[:temp - 1] != str(res)[:temp - 1]:
			temp_res = int(str(res)) + 1
			while str(temp_res) != str(temp_res)[::-1]:
				temp_res += 1
			b -= 1
			if b >= 0:
				l.append(temp_res)
		res += int(step)
	return l[-1]


def palin(a, b):
	base = str(10 ** ((a - 1) // 2) + b - 1)
	return int(base + base[::-1][a % 2:])


def palin(a, b):
	n = (a - 1) // 2
	(x, y) = divmod(b - 1, 10 ** n)
	result = str(x + 1) + (a > 2) * f'{y:0{n}}'
	return int(result + result[-1 + a % -2::-1])


def palin(n, pos):
	half = str(int('1' + '0' * (n // 2 - int(not n & 1))) + pos - 1)
	full = int(half + [half, half[:-1]][n & 1][::-1])
	return full


from math import ceil

def palin(a, b):
	c = ceil(a / 2)
	for i in range(10 ** (c - 1), 10 ** c):
		p = str(i)
		p += p[::-1][a % 2:]
		if b == 1:
			return int(p)
		else:
			b -= 1


def palin(a, b):
	left = str(10 ** ((a - 1) // 2) + (b - 1))
	right = str(left[::-1][a % 2:])
	return int(left + right)