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:
A = ['w', 'W', 'T', 'Y', 'U', 'I', 'o', 'O', 'A', 'H', 'X', 'x', 'V', 'v', 'M', 'b', 'd', 'p', 'q']
s = input()
f = True
for i in range(len(s)):
	if s[i] not in A:
		f = False
	if s[i] == "b" and s[-i - 1] != "d":
		f = False
	if s[i] == "d" and s[-i - 1] != "b":
		f = False
	if s[i] == "p" and s[-i - 1] != "q":
		f = False
	if s[i] == "q" and s[-i - 1] != "p":
		f = False
	if s[i] not in ["p", "q", "b", "d"] and s[i] != s[-i - 1]:
		f = False
if f:
	print("TAK")
else:
	print("NIE")