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()
a = 'aBCcDEeFfGghiJjKkLlmNnPQRrSstuyZz'
b = 'AbdHIMOopqTUVvWwXxY'
c = 'AdbHIMOoqpTUVvWwXxY'
f = True
n = len(s)
for i in range(0, n):
  if s[i] in a:
    f = False
  elif c[b.index(s[i])] != s[n-i-1]:
    f = False
  if f == False:
    break
print('TAK' if f else 'NIE')