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:
def mirror(text):
    for i in range(len(text)):
        if text[i] not in 'AHIMOoTUVvWwXxY':
            try:
                rules = {'b': 'd', 'd': 'b', 'p': 'q', 'q': 'p'}
                if rules[text[i]] != text[-i-1]:
                    return 'NIE'
            except Exception:
                return 'NIE'
        else:
            if text[i] != text[-i-1]:
                return 'NIE'
    return 'TAK'


print(mirror(input()))