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:
insym = set(('A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y'))
disym = {'b':'d', 'd':'b', 'p':'q', 'q':'p'}
for c in insym:
    disym[c] = c
s = input()
n = len(s)
if n%2==0 or s[n//2] in insym:
    s1 = s[0:n//2]
    s2 = s[::-1][0:n//2]
    flg = True
    for i in range(n//2):
        if s1[i] not in disym or disym[s1[i]] != s2[i]:
            flg = False
    if flg:
        print("TAK")
    else:
        print("NIE")
else:
    print("NIE")