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