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 Main {
	
	public static void main(String [] arge)
	{
		String s="AbdHIMOopqTUVvWwXxY",ss="AdbHIMOoqpTUVvWwXxY";
		char key[] = new char[300];
		for(int i=0;i<19;i++) key[s.charAt(i)]=ss.charAt(i);
		String s1;
		Scanner cin = new Scanner(System.in);
		s1=cin.nextLine();
		boolean flag=true;
		for(int i=0;i<s1.length();i++)
		{
			if(s1.charAt(s1.length()-1-i)!=key[s1.charAt(i)])
			 flag=false;
			
		}
		if(flag) System.out.println("TAK");
		else System.out.println("NIE");
		
		
	}
	

}