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.HashMap;
import java.util.HashSet;
import java.util.Hashtable;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		char [] c={};
		String st,s= "AHIMOoTUVvWwXxYbdpq"
				,fl="AHIMOoTUVvWwXxYdbqp";
		HashMap<Character, Character> hm= new HashMap<>();
		int i;
		for(i=0;i<s.length();i++){
			hm.put(s.charAt(i), fl.charAt(i));
		}
		Scanner sc= new Scanner(System.in);
		st=sc.nextLine();
		for(i=0;i<((st.length()%2==0)?st.length()/2:st.length()/2+1);i++){
			if(!hm.containsKey(st.charAt(i) ))
			{
				System.out.println("NIE");return;
			}
			else if (hm.get(st.charAt(i) )!= st.charAt(st.length()-1-i)){
				System.out.println("NIE");return;
			}
		}
		System.out.println("TAK");
	}

}