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()
D = {'A': 'A', 'b': 'd', 'd': 'b', 'H': 'H', 'I': 'I', 'M': 'M', 'O': 'O', 'o': 'o', 'p': 'q', 'q': 'p', 'T': 'T', 'U': 'U', 'V': 'V', 'v': 'v', 'W': 'W', 'w': 'w', 'X': 'X', 'x': 'x', 'Y': 'Y'}
for (c1, c2) in zip(s, s[::-1]):
    if D.get(c1, '') != c2:
        print("NIE")
        exit()
print("TAK")