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.*;

public class ec14b {

	public static boolean isvalid(char c) {
		boolean b = false;
		if (c == 'o' || c == 'A' || c == 'H' || c == 'I' || c == 'M' || c == 'O' || c == 'T' || c == 'U' || c == 'V'
				|| c == 'W' || c == 'X' || c == 'Y' || c == 'v' || c == 'w' || c == 'x') {
			b = true;
		}
		return b;
	}

	public static void main(String args[]) {
		Scanner scan = new Scanner(System.in);
		String s = scan.next();
		StringBuilder sb = new StringBuilder(s);
		String rev = sb.reverse().toString();
		for (int i = 0; i < s.length(); i++) {
			if (!(isvalid(s.charAt(i)) && isvalid(rev.charAt(i))) || s.charAt(i) != rev.charAt(i)) {
				if ((s.charAt(i) == 'b' && rev.charAt(i) == 'd') || (s.charAt(i) == 'd' && rev.charAt(i) == 'b')
						|| (s.charAt(i) == 'p' && rev.charAt(i) == 'q')
						|| (s.charAt(i) == 'q' && rev.charAt(i) == 'p')) {
					continue;
				}
				System.out.println("NIE");
				return;
			}
		}
		System.out.println("TAK");
	}
}