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:
a={"A":"A","H":"H","I":"I","M":"M","O":"O","T":"T","U":"U","V":"V","W":"W","X":"X","Y":"Y","b":"d","d":"b",
   "o":"o","v":"v","w":"w","x":"x","q":"p","p":"q"}
s=input().rstrip()
n=len(s)
ans="TAK"
for i in range(n//2):
    if a.get(s[i])==None or a[s[i]]!=s[-(i+1)]:
        ans="NIE"
        break
if n%2 and (a.get(s[n//2])==None or a[s[n//2]]!=s[n//2]):
    ans="NIE"
print(ans)