Problem: Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.

<image> English alphabet

You are given a string s. Check if the string is "s-palindrome".

Input

The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters.

Output

Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise.

Examples

Input

oXoxoXo


Output

TAK


Input

bod


Output

TAK


Input

ER


Output

NIE

Solution (in PYTHON3):

Solution:
s = input()
#print(s)
mir = 'xXAoOWwTYIMUvVH'

def ch(a, b, x): 
	return (s[x] == 'b' and s[-x - 1] == 'd') or (s[x] == 'p' and s[-x - 1] == 'q') or (s[x] == 'd' and s[-x - 1] == 'b') or (s[x] == 'q' and s[-x - 1] == 'p')

def ispal(s):
	flag = True
	for x in range(len(s)//2):
		if ((s[x] == s[-x - 1] and (s[x] in mir)) or ch(s[x], s[-x - 1], x)) and (len(s) % 2 == 0 or s[len(s)//2] in mir): pass 
		else: flag = False
	return flag

if len(s) == 1 and s[0] in mir: print('TAK')
elif len(s) != 1 and ispal(s): print('TAK')
else: print('NIE')