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:
def main():
	s = input()
	print(solver(s))

def solver(s):
	mirrors = {'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 i in range((len(s) + 1) // 2):
		left = s[i]
		right = s[len(s) - i - 1]
		if left not in mirrors or right != mirrors[left]:
			return "NIE"
	return "TAK"

# print(solver("oXoxoXo"))
# print(solver("bod"))
# print(solver("ER"))

main()