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:
s = input()
h = ['A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'v', 'V', 'W', 'w', 'X', 'x', 'Y']
q = ['A', 'b', 'd', 'H', 'I', 'M', 'O', 'o', 'p', 'q', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']
w = ['A', 'd', 'b', 'H', 'I', 'M', 'O', 'o', 'q', 'p', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']
g = False
if len(s) % 2:
    x = s[(len(s) + 1) // 2:]
    z = s[:(len(s) - 1) // 2]
    a = s[(len(s) - 1) // 2]
    z = list(reversed(z))
    if a not in h:
        g = True
    for i in range(len(z)):
        if g:
            break
        if x[i] not in q:
            g = True
            break
        if w[q.index(x[i])] != z[i]:
            g = True
    if g:
        print("NIE")
    else:
        print("TAK")
else:
    x = s[(len(s)) // 2:]
    z = s[:(len(s)) // 2]
    z = list(reversed(z))
    for i in range(len(z)):
        if g:
            break
        if x[i] not in q:
            g = True
            break
        if w[q.index(x[i])] != z[i]:
            g = True
    if g:
        print("NIE")
    else:
        print("TAK")