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:

# coding: utf-8

# In[12]:

SLETTERS = (('A', 'A'), ('b', 'd'), ('H', 'H'), ('I', 'I'),
            ('M', 'M'), ('O', 'O'), ('o', 'o'), ('p', 'q'),
            ('T', 'T'), ('U', 'U'), ('V', 'V'), ('v', 'v'),
            ('W', 'W'), ('w', 'w'), ('X', 'X'), ('x', 'x'),
            ('Y', 'Y'))

LETTERS = 'AHIMOoTUVvWwXxY'

s = input()
n = len(s)

if n == 1:
    s_pal = 'TAK' if s in LETTERS else 'NIE'
else:
    s_pal = 'TAK'
    for i in range(n // 2):
        if (s[i], s[n - i - 1]) not in SLETTERS and (s[n - i - 1], s[i]) not in SLETTERS:
            s_pal = 'NIE'

    if n % 2 == 1 and s[n // 2] not in LETTERS:
        s_pal = 'NIE'
        
print(s_pal)


# In[ ]: