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:
g={}
g['A']='A'
g['H']='H'
g['I']='I'
g['M']='M'
g['O']='O'
g['T']='T'
g['U']='U'
g['V']='V'
g['W']='W'
g['X']='X'
g['Y']='Y'
g['b']='d'
g['d']='b'
g['p']='q'
g['q']='p'
g['o']='o'
g['v']='v'
g['w']='w'
g['x']='x'
a=0
s=list(raw_input())
count,ans=0,0
for i in range(len(s)):
	if s[i] in g: count+=1
if count!=len(s): print 'NIE'
else:
	for i in range(len(s)):
		if s[i]==g[s[len(s)-1-i]]: ans+=1
	if ans==len(s): print 'TAK'
	else: print 'NIE'