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:
self_sym_list = ['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']
#has_no_sym_list = ['a','B','C','c','D','E','e','F','f','G','g','h','i']
sym_dict = {'b': 'd', 'd': 'b', 'p': 'q', 'q': 'p'}

sym_dict.update( {k: k for k in self_sym_list} )
sym_list = self_sym_list + list(sym_dict.keys())

ans = True
ind = 0
s = input()
if len(s) % 2 == 1:
    ind = len(s) // 2
    if s[ind] not in self_sym_list:
        ans=False
    ind -= 1
else:
    ind = len(s) // 2 - 1
    
for i in range(ind+1):
    if (s[i] not in sym_dict) or (sym_dict[s[i]] != s[len(s)-1-i]):
        ans=False
        break

if ans:
    print('TAK')
else:
    print('NIE')