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:

s = input()


d = {
    'A': 'A',
    'b': 'd',
    'd': 'b',
    'H': 'H',
    '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'
}

is_sp = True
length = len(s)

left_i = 0
right_i = length - 1
while left_i <= right_i:
    left = s[left_i]
    right = s[right_i]
    if left not in d or d[left] != right:
        is_sp = False
        break
    left_i += 1
    right_i -= 1

if is_sp:
    print("TAK")
else:
    print("NIE")