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:
/*input
xox
*/
import java.io.*;
import java.util.*;


public class Main {
    public static void main(String []args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        String line = br.readLine();
        Map<Character, Character> map = new HashMap<>();
        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');

        char[] arr = new char[line.length()];
        for (int i = 0; i < arr.length; i++) {
            Character c = map.get(line.charAt(i));
            arr[i] = (c == null) ? ' ' : c;
        }
        String rev = new String(arr);
        rev = new StringBuilder(rev).reverse().toString();
        System.out.println(rev.equals(line) ? "TAK" : "NIE");
    }
}