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:
z='AHIMOoTUVvWwXxY'
t=[('b','d'),('q','p')]
d={}
for i,j in t:d[i]=j;d[j]=i
for i in z:d[i]=i
s=input()
n=len(s)
a='TAK'
if n%2 and s[n//2] not in z:a='NIE'
for i in range(n//2):
 if d.get(s[i],0)!=s[n-1-i]:a='NIE'
print(a)