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()
n = len(s)
c = ['A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']
p = ['b', 'd', 'q', 'p']

ok = 1
for i in range(n):
	if s[i] == s[n-i-1] or (s[i] in p and s[n-i-1] in p):
		continue
	ok = 0
if ok == 1:
	for i in range(n):
		if s[i] in c:
			continue
		elif s[i] in p:
			if s[i] == p[0] and s[n-i-1] == p[1]:
				continue
			elif s[i] == p[1] and s[n-i-1] == p[0]:
				continue
			elif s[i] == p[2] and s[n-i-1] == p[3]:
				continue
			elif s[i] == p[3] and s[n-i-1] == p[2]:
				continue
		ok = 0

if ok == 1:
	print("TAK")
else:
	print("NIE")