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:
# in the name of god
import re

string = str(input())
count = 0
s = ("A", "O", "W", "Y", "T", "X", "o", "w", "x", "I", "M", "U", "V", "v", "H")
for i in range(0, int(len(string) / 2)):
    if (string[i] == string[len(string) - i - 1] and string[i] in s) or (
            string[i] == "b" and string[len(string) - i - 1] == "d") or (
            string[i] == "d" and string[len(string) - i - 1] == "b") or (
            string[i] == "p" and string[len(string) - i - 1] == "q") or (
            string[i] == "q" and string[len(string) - i - 1] == "p"):
        count += 1
if len(string) % 2 == 1:
    if string[int(len(string) / 2)] not in s:
        count -= 1
if count == int(len(string) / 2):
    print("TAK")
else:
    print("NIE")