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 = {
  'A' : 'A',
  'b' : 'd',
  'd' : 'b',
  'H' : 'H',
  'I' : 'I',
  'M' : 'M',
  'O' : 'O',
  'o' : 'o',
  'X' : 'X',
  'x' : 'x',
  'Y' : 'Y',
  'W' : 'W',
  'V' : 'V',
  'w' : 'w',
  'v' : 'v',
  'T' : 'T',
  'p' : 'q',
  'q' : 'p',
  'U' : 'U'
  }
g = lambda c : '*' if c not in d.keys() else d[c]
s = input()
for i in range(len(s)) :
  if s[i] != g(s[len(s)-i-1]) :
    print('NIE')
    exit(0)
print('TAK')