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:
__author__ = 'Alexander'
import sys
string = sys.stdin.readline().strip()
ideal = {'A','b','d','H','I','M','O','o','p','q','T','U','V','v','W','w','X','x','Y'}

for i in range(int((len(string)+1)/2)):
    if string[i] not in ideal:
        sys.stdout.write("NIE")
        sys.exit()
    elif string[i] == 'b' or string[i] == 'd' or string[i] == 'p' or string[i] == 'q':
        if (string[i] == 'b' and string[-i-1] != 'd') or \
        (string[i] == 'd' and string[-i-1] != 'b') or \
        (string[i] == 'q' and string[-i-1] != 'p') or \
        (string[i] == 'p' and string[-i-1] != 'q'):
            sys.stdout.write("NIE")
            sys.exit()
    elif string[i] != string[-i-1]:
        sys.stdout.write("NIE")
        sys.exit()
sys.stdout.write("TAK")