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:
m = {'b':'d','d':'b','p':'q','q':'p'}
a = 'AHIMOTUVWXYovwx'
f = lambda c :m[c] if c in m.keys() else c if a.count(c)  else '#'
s = input()
for i in range(len(s)):
  if s[i] != f(s[len(s) - i - 1]):
    print('NIE')
    exit(0)
print('TAK')