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 {
	static Scanner sc = new Scanner (System.in);
	static int []arr = new int [1005];
	static char s[] = {'A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y'};
	static String ss;
	public static void main(String[]args)
	{
		ss = sc.next();
		int len = ss.length();
		int i = 0 , j = len - 1;
		boolean flag = false;
		while(i <= j)
		{
			if(check(i,j))
			{
				i ++;
				j --;
			}else 
			{
				System.out.println("NIE");
				System.exit(0);
			}
			
		}
		System.out.println("TAK");
	}
	private static boolean check(int i, int j) {
		if(ss.charAt(i) == 'b' && ss.charAt(j) == 'd' 
		|| ss.charAt(i) == 'd' && ss.charAt(j) == 'b'
		|| ss.charAt(i) == 'p' && ss.charAt(j) == 'q'
		|| ss.charAt(i) == 'q' && ss.charAt(j) == 'p')
			return true;
		if(ss.charAt(i)  == ss.charAt(j))
		{
			for(int k = 0 ; k < s.length; k ++)
			{
				if(ss.charAt(i) == s[k])
					return true;
			}
		}
		return false;
	}
}