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)

N = 0

if n % 2 == 0:
	N =  n // 2
else:
	N = n // 2 + 1
flag = True

def check(l, r):
	same = ["A", "H", "I", "M", "O", "o", "T", "U", "V", "v", "W", "w", "X", "x", "Y"]
	if (l == r) and (l in same):
		return True
	if l == "b" and r =="d":
		return True
	if l == "d" and r == "b":
		return True
	if l == "p" and r == "q":
		return True
	if l == "q" and r == "p":
		return True
	return False
for i in range(N):
	left = i
	right = n-1-i
	lc = s[left]
	rc = s[right]

	if ( not check(lc, rc) ):
		flag = False

if flag:
	print("TAK")
else:
	print("NIE")