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()
half1 = s[:(len(s)+1)//2]
half2 = s[len(s)//2:]

r = {
    'a': '#',
    'b': 'd',
    'c': '#',
    'd': 'b',
    'e': '#',
    'f': '#',
    'g': '#',
    'h': '#',
    'i': '#',
    'j': '#',
    'k': '#',
    'l': '#',
    'm': '#',
    'n': '#',
    'o': 'o',
    'p': 'q',
    'q': 'p',
    'r': '#',
    's': '#',
    't': '#',
    'u': '#',
    'v': 'v',
    'w': 'w',
    'x': 'x',
    'y': '#',
    'z': '#',
    'A': 'A',
    'B': '#',
    'C': '#',
    'D': '#',
    'E': '#',
    'F': '#',
    'G': '#',
    'H': 'H',
    'I': 'I',
    'J': '#',
    'K': '#',
    'L': '#',
    'M': 'M',
    'N': '#',
    'O': 'O',
    'P': '#',
    'Q': '#',
    'R': '#',
    'S': '#',
    'T': 'T',
    'U': 'U',
    'V': 'V',
    'W': 'W',
    'X': 'X',
    'Y': 'Y',
    'Z': '#'
}

half2bis = ""
for l in half2:
    half2bis += r[l]

print("TAK" if half1 == half2bis[::-1] else "NIE")