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:
string = input()
dic = {"o": "o", "x": "x", "d": "b", "b": "d", "v": "v", "w": "w", "p": "q", "q": "p"}
mid = len(string) // 2 + 1
is_p = True
for i in range(1, mid):
    f, s = string[i - 1], string[i * -1]
    if (f == s and f in "AHIMOTUVWXY") or (f in dic and dic[f] == s):
        continue
    else:
        is_p = False
        break
if len(string) % 2 == 1:
    if string[mid - 1] not in "AHIMOTUVWXYoxvw":
        is_p = False
if len(string) == 1:
    if (string in "AHIMOTUVWXY") or (string in dic and dic[string] == string):
        print("TAK")
    else:
        print("NIE")
elif is_p:
    print("TAK")
else:
    print("NIE")