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:
s=list(raw_input())
l=s[:(len(s)+1)/2]
r=s[len(s)/2:];r.reverse()
j=['A','b','d','H','I','M','O','o','p','q','T','U','V','v','W','w','X','x','Y']
k=['A','d','b','H','I','M','O','o','q','p','T','U','V','v','W','w','X','x','Y']
for i in range(len(r)):
    if not (r[i] in j):
        print 'NIE'
        quit()
    r[i]=k[j.index(r[i])]
print ['NIE','TAK'][l==r]