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


s=raw_input().strip()
n=len(s)
for i in xrange(n):
    
    if (s[n-1-i] not in d or s[i]!=d[s[n-1-i]]):
        print "NIE"
        break
else:
    print "TAK"