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.Map;
import java.util.Scanner;

public class Main {
    public static void main(String [] argv){
        Scanner in = new Scanner(System.in);
        String s;
        String str[] = {"A","b","d","H","I","M","O","o","p","q","T","U","V","v","W","w","X","x","Y"};
        Map<String,String> map=new HashMap<String,String>();
        map.clear();
        map.put("A","A");
        map.put("b","d");
        map.put("d","b");
        map.put("H","H");
        map.put("I","I");
        map.put("M","M");
        map.put("O","O");
        map.put("o","o");
        map.put("p","q");
        map.put("q","p");
        map.put("T","T");
        map.put("U","U");
        map.put("V","V");
        map.put("v","v");
        map.put("W","W");
        map.put("w","w");
        map.put("X","X");
        map.put("x","x");
        map.put("Y","Y");
        while(in.hasNext())
        {
            s = in.next();
            int j;
            boolean f = true;
            for(int i = 0; i < (s.length() + 1) / 2; ++i)
            {
                for(j = 0; j < 19; ++j)
                {
                    if (str[j].equals(String.valueOf(s.charAt(i))))
                        break;
                }
                if(j == 19)
                {
                    f = false;
                    break;
                }
                if(String.valueOf(s.charAt(s.length() - 1 - i)).equals(map.get(String.valueOf(s.charAt(i)))))
                {
                    continue;
                }
                else
                {
                    f = false;
                    break;
                }
            }
            if(f == true)
                System.out.println("TAK");
            else
                System.out.println("NIE");
        }
        in.close();
    }
}