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:
from sys import exit

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

half_sym = ['A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y',
            'o', 'v', 'w', 'x']

d = {}
for t in sym:
    a, b = t
    d[a], d[b] = b, a
    
s = input()
if len(s) % 2 == 1:
    c = len(s) // 2
    if s[c] not in half_sym:
        print('NIE')
        exit(0)
    l = c - 1
    r = c + 1
else:
    r = len(s) // 2
    l = r - 1

while l >= 0:
    if s[l] not in d or d[s[l]] != s[r]:
        print('NIE')
        exit(0)
    l -= 1
    r += 1
print('TAK')