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=input()
fix={'A':'A',
     'B':'.',
     'C':'.',
     'D':'.',
     'E':'.',
     'F':'.',
     'G':'.',
     'H':'H',
     'I':'I',
     'J':'.',
     'K':'.',
     'L':'.',
     'M':'M',
     'N':'.',
     'O':'O',
     'P':'.',
     'Q':'.',
     'R':'.',
     'S':'.',
     'T':'T',
     'U':'U',
     'V':'V',
     'W':'W',
     'X':'X',
     'Y':'Y',
     'Z':'.',
     'a':'.',
     'b':'d',
     'c':'.',
     'd':'b',
     'e':'.',
     'f':'.',
     'g':'.',
     'h':'.',
     'i':'.',
     'j':'.',
     'k':'.',
     'l':'.',
     'm':'.',
     'n':'.',
     'o':'o',
     'p':'q',
     'q':'p',
     'r':'.',
     's':'.',
     't':'.',
     'u':'.',
     'v':'v',
     'w':'w',
     'x':'x',
     'y':'.',
     'z':'.'}

for i in range(len(a)//2):
    if fix[a[i]]==a[len(a)-i-1]:
        pass
    else:
        print('NIE')
        quit()
if len(a)%2==1 and fix[a[len(a)//2]]!=a[len(a)//2]:
    print('NIE')
    quit()
print('TAK')