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()
l=len(s)
li=['A','O','o','H','V','v','X','x','W','w','M','T','I','U','Y']
li1=['pq','qp','bd','db']
if(l&1):
	if s[l//2] not in li:
		print("NIE")
		exit()
f=0
for i in range(len(s)//2):
	if(s[i]==s[l-1-i]):
		if s[i] not in li:
			f=1
			break
	else:
		if s[i]+s[l-1-i] not in li1:
			f=1
			break
			
if(f):
	print("NIE")
else:
	print("TAK")