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:
p={'b':'d','d':'b','p':'q','q':'p'}
#c=['A','H','I','i','l','M','O','o','T','U','V','v','W','w','X','x','Y']
c=['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']
s=raw_input()
if len(s)%2==1:
    if s[len(s)/2] not in c:
        print 'NIE'
        exit()
for i in range(len(s)):
    if s[i] in c and s[-1*i-1]==s[i]:
        pass
    elif s[i] in p and p[s[i]]==s[-1*i-1]:
        pass
    else:
        print 'NIE'
        exit()
print 'TAK'