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 = str(input())

n = len(s)

ss = ['A', 'H', 'I', 'M', 'O','o','T','U','V','v','W','w','X','x','Y']
ss2 = [('d','b'),('q','p'),('p','q'),('b','d')]

for i in range(0, n//2):
    if s[i]==s[n-1-i]:
        if not (s[i] in ss):
            print("NIE")
            exit()
    else:
        if not ((s[i],s[n-1-i]) in ss2):
            print("NIE")
            exit()

if n%2 != 0:
    if s[n//2] in ss:
        print("TAK")
    else:
        print("NIE")
else:
    print("TAK")