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 PYTHON):

Solution:
import sys
s = raw_input()

fine = ['A','H','I','M','O','o','T','U','V','v','W','w','x','X','Y']
check = ['p','q','b','d']

for i in xrange(len(s)):
	if s[i] in fine:
		pass
	elif s[i] in check:
		if s[i] == 'p' and s[len(s)-1-i]=='q':
			pass
		elif s[i] == 'q' and s[len(s)-1-i]=='p':
			pass
		elif s[i] == 'd' and s[len(s)-1-i]=='b':
			pass
		elif s[i] == 'b' and s[len(s)-1-i]=='d':
			pass
		else:
			print 'NIE'
			sys.exit(0)
	else:
		print 'NIE'
		sys.exit(0)
s = s.replace('p','q')
s = s.replace('b','d')

if s == s[::-1]:
	print 'TAK'
else:
	print 'NIE'