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()
good = ['A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']
for i in range(len(s) // 2):
    if (s[i] == s[len(s) - i - 1] and s[i] in good) or (s[i] == 'b' and s[len(s) - i - 1] == 'd') or (s[i] == 'p' and s[len(s) - i - 1] == 'q') or (s[i] == 'd' and s[len(s) - i - 1] == 'b') or (s[i] == 'q' and s[len(s) - i - 1] == 'p'):
        pass
    else:
        print("NIE")
        exit()
if len(s) % 2 == 1:
    if s[len(s) // 2] in good:
        print("TAK")
    else:
        print("NIE")
else:
    print("TAK")