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()
itself = {'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y'}
d = {'p' : 'q', 'q' : 'p', 'b' : 'd', 'd' : 'b'}
for i in itself:
	d[i] = i
ok = True
for i in range(len(s)):
	ok &= s[i] in d.keys() and s[len(s) - i - 1] == d[s[i]]
print("TAK" if ok else "NIE")