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:
from sys import stdin

mot=stdin.readline()
longueur=len(mot)-1
milieu=longueur//2
couples=[("A","A"),("H","H"),("I","I"),("M","M"),("O","O"),("o","o"),("T","T"),("U","U"),("V","V"),("v","v"),("W","W"),("w","w"),("X","X"),("x","x"),("Y","Y"),("b","d"),("d","b"),("p","q"),("q","p")]

inf=milieu
sup=milieu
if longueur%2==0:
	inf-=1

while inf>=0 and sup<longueur and (mot[inf],mot[sup]) in couples:
	inf-=1
	sup+=1

if sup>=longueur:
	print("TAK")
else:
	print("NIE")