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:
import sys
s = input()
n = len(s)
m = {'A' : 'A', 'b' : 'd', 'd' : 'b', 'H' : 'H', 'I' : 'I', 'M' : 'M', 'O' : 'O', 'o' : 'o', 'p' : 'q', 'q' : 'p', 'T' : 'T', 'U' : 'U', 'V' : 'V', 'v' : 'v', 'W' : 'W', 'w' : 'w', 'X' : 'X', 'x' : 'x', 'Y' : 'Y'}
for i in range(0, n // 2):
    if not (s[i] in m) or s[n - 1 - i] != m[s[i]]:
        print ("NIE")
        sys.exit()
if n%2:
    i = n//2
    if not s[i] in ['A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y']:
        print ("NIE")
        sys.exit()
print ("TAK")