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 PYTHON):

Solution:
left = ['b', 'd', 'p', 'q', 'A','H', 'I', 'M', 'O', 'o', \
			'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']
right = ['d', 'b', 'q', 'p', 'A', 'H', 'I', 'M', 'O', 'o', \
			'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']
s = raw_input()
def check(s):
	for i in xrange(len(s)/2):
		if s[i] not in set(left) or s[len(s)-1-i] not in set(left):
			return False
		if left.index(s[i]) != right.index(s[len(s)-1-i]):
			return False
	if len(s)%2 == 0:
		return True
	elif s[len(s)/2] in set(left[4:]):
		return True
	else:
		return False

if check(s):
	print 'TAK'
else:
	print 'NIE'