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:
text = input()
if len(text) % 2 == 0:
    example = text[0:(len(text) // 2)]
    ex = text[(len(text) // 2):]
else:
    example = text[0:(len(text) // 2) + 1]
    ex = text[(len(text) // 2):]
ex = ex[::-1]
ex2 = ""
for i in ex:
    p = ["o","v","w","x","A","H","I","M","O","T","U","V","W","X","Y"]
    if i == "b":
        ex2 += "d"
    elif i == "d":
        ex2 += "b"
    elif i == "p":
        ex2 += "q"
    elif i == "q":
        ex2 += "p"
    elif p.count(i) == 1:
        ex2 += i
    else: ex2 += "0"
if example == ex2:
    print("TAK")
else: print("NIE")