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:
d=['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']
s=input()
l=0
r=len(s)-1
f=1
while(l<=r):
    if (s[l] in d and s[r] in d and s[l]==s[r]) or (s[l]=='b' and s[r]=='d') or (s[l]=='d' and s[r]=='b') or (s[l]=='p' and s[r]=='q') or (s[l]=='q' and s[r]=='p'):
        l+=1
        r-=1
    else:
        f=0
        break
if f:
    print('TAK')
else:
    print('NIE')