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()
d = {"b" : "d","d":"b","o":"o","p":"q","q":"p","v":"v","w":"w","x":"x","A":"A","H":"H","I":"I","M":"M","O":"O","T":"T","U":"U","V":"V","W":"W","X":"X","Y":"Y"}
n = len(s)//2
flag = 0
l = []
if len(s)%2 == 0:
    i = n-1
    j = n

else:
    i = n
    j = n

while i > -1:
    if s[i] not in d:
        flag = 1
        break

    l.append(d[s[i]])
    i -= 1

if flag:
    print("NIE")
    exit()

o = []
for e in range(j,len(s)):
    o.append(s[e])

if o == l:
    print("TAK")

else:
    print("NIE")