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:
d = {'o': 'o', 'I': 'I', 'H': 'H', 'b': 'd', 'x': 'x',
     'X': 'X', 'V': 'V', 'v': 'v', 'U': 'U', 'W': 'W', 'w': 'w',
     'M': 'M', 'A': 'A', 'Y': 'Y', 'T': 'T', 'p': 'q', 'q': 'p',
     'd': 'b', 'O': 'O'}

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

s = input()

if len(s) % 2 == 0:
    n = len(s) // 2
    j = len(s) // 2 + 1
    i = 0
    while i < n:
        if s[i] not in d:
            print('NIE')
            exit()
        if d[s[i]] != s[len(s) - i - 1]:
            print('NIE')
            exit()
        i += 1
    print('TAK')
else:
    if s[len(s) // 2] not in c:
        print('NIE')
        exit()
    for i in range(len(s) // 2):
        if s[i] not in d:
            print('NIE')
            exit()
        if d[s[i]] != s[len(s) - i - 1]:
            print('NIE')
            exit()

    print('TAK')