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:
inv={'d':'b','b':'d','A':'A','O':'O','o':'o','p':'q','q':'p','x':'x','X':'X','Y':'Y','w':'w','W':'W','v':'v','V':'V','U':'U','T':'T','M':'M','I':'I','H':'H'}

def is_sym(s):
    l=len(s)
    for c in range(l//2+1):
        if s[c] in inv:
            if inv[s[c]]!=s[l-c-1]:
                return 0
        else:
            return 0
    return 1

s=input()
if is_sym(s):
    print('TAK')
else:
    print('NIE')