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:
d={'A':'A','b':'d','d':'b','H':'H','I':'I','M':'M','O':'O','o':'o','p':'q','q':'p','T':'T','U':'U','V':'V','v':'v','W':'W','w':'w','X':'X','x':'x','Y':'Y'}
s=raw_input()
l=len(s)
i=0
j=l-1
yes=True
while i<=j:
	if s[i] not in d.keys() :
		yes=False
		break
	elif d[s[i]]!=s[j] :
		yes=False
		break
	i+=1
	j-=1
if yes :
	print "TAK"
else :
	print "NIE"