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:
def main():
    check = list('AbdHIMOopqTUVvWwXxY')

    qsn = list(input())

    test = [x for x in qsn if x in check]

    if test != qsn:
        print("NIE")
        return
    dict = {'Y': 'Y', 'v': 'v', 'V': 'V', 'W': 'W', 'H': 'H', 'O': 'O', 'A': 'A', 'w': 'w', 'b': 'd', 'x': 'x', 'q': 'p', 'p': 'q', 'o': 'o', 'M': 'M', 'T': 'T', 'd': 'b', 'I': 'I', 'X': 'X', 'U': 'U'}

    length = len(qsn)

    temp1 = qsn[:int(length/2)]
    temp2 = qsn[:-(int((length/2))+1):-1]
    for i in range(int(length/2)):
        if dict[temp1[i]] == temp2[i]:
                flag = 1
        else:
                print("NIE")
                return
    if(length%2 != 0 and qsn[int(length/2)] not in list('AHIMOoTUVWwXxY')):
        print("NIE")
        return


    print("TAK")
    return

main()