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 main():
    non_symetrical = {'b', 'd', 'p', 'q'}
    chars = {
        'A': 'A',
        'H': 'H',
        'b': 'd',
        'd': 'b',
        'I': 'I',
        'M': 'M',
        'O': 'O',
        'o': 'o',
        'p': 'q',
        'q': 'p',
        'T': 'T',
        'U': 'U',
        'V': 'V',
        'v': 'v',
        'W': 'W',
        'w': 'w',
        'X': 'X',
        'x': 'x',
        'Y': 'Y',
    }
    alpha = set(chars.keys())
    s = input()
    n = len(s)
    if n % 2 == 1:
        middle = s[n//2]
        if middle not in alpha or middle in non_symetrical:
            print('NIE')
            return

    for i in range(n//2+1):
        first, last = s[i], s[n - i - 1]
        if first not in alpha or last not in alpha:
            print('NIE')
            return
        if chars[first] != last:
            print('NIE')
            return

    print('TAK')

if __name__ == '__main__':
    main()