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()

mirror={'b':'d','d':'b','p':'q','q':'p'}
while len(s)>1:
        if s[0] in 'AoOIMHTUVvWwXxY':
                if s[0]==s[-1]: s=s[1:-1:]
                else: break
        elif s[0] in mirror:
                if s[0]==mirror[s[-1]]: s=s[1:-1:]
                else: break
        else: break

if len(s)==0 or (s[0] in 'AoOIMHTUVvWwXxY' and len(s)==1):
        print('TAK')
else: print('NIE')