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.io.*;
import java.util.*;
import java.math.*;

public class e14B  {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		char[] s = in.next().toCharArray();
		int ind = 0;
		while(ind<=s.length-ind-1)
		{
			char a = s[ind];
			char b = s[s.length-ind-1];
			
			char expected = '\0';
			switch(a)
			{
				case 'b':
					expected = 'd';
					break;
				case 'd':
					expected = 'b';
					break;
				case 'p':
					expected = 'q';
					break;
				case 'q':
					expected = 'p';
					break;
				case 'A':
				case 'H':
				case 'I':
				case 'M':
				case 'O':
				case 'o':
				case 'T':
				case 'U':
				case 'V':
				case 'v':
				case 'W':
				case 'w':
				case 'X':
				case 'x':
				case 'Y':
					expected = a;
					break;
				default:
					expected = '\0';
					break;
			}
			if(expected!=b)
			{
				System.out.println("NIE");
				return;
			}
			ind++;
		}
		System.out.println("TAK");
	}
}