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 JAVA):

Solution:
import java.util.Scanner;
public class Sol {
	public static void main(String args[]) 
	{ 
		java.util.Scanner in = new Scanner(System.in);
		String palindrome = "Vv Ww Oo Xx M A H I U Y T";
		String s = in.next();
		for(int i=0;i<s.length();i++) {
			if((palindrome.contains(s.charAt(i)+"") && ((s.charAt(i) == s.charAt(s.length()-i-1))) || (s.charAt(i)=='p' && s.charAt(s.length()-i-1)=='q')||(s.charAt(i)=='q' && s.charAt(s.length()-i-1)=='p')||(s.charAt(i)=='d' && s.charAt(s.length()-i-1)=='b')||(s.charAt(i)=='b' && s.charAt(s.length()-i-1)=='d')))
				continue;
			else {
				System.out.println("NIE");
				return ;
			}
		}
		System.out.println("TAK");
	}
}