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:
lst = [
    'AA', 'HH', 'II', 'MM', 'OO', 'TT', 'UU', 'VV', 'WW', 'XX', 'YY',
    'bd', 'db', 'oo', 'pq', 'qp', 'vv', 'ww', 'xx'
]


def palindrom(s):
    l, r = 0, len(s) - 1
    while l <= r:
        if s[l] + s[r] not in lst:
            return "NIE"
        l += 1
        r -= 1
    return "TAK"


print(palindrom(input()))