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[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine();
		boolean flag = false;
		boolean found;
		char[] set = {'A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y'};
		int last = s.length() - 1;
		if (s.length() % 2 != 0) {
			int middle = s.length() / 2;
			for (int i = 0; i < set.length && !flag; i++) {
				if (s.charAt(middle) == set[i]) {
					flag = true;
				}
			}
			if (flag == true) {
				flag = false;
			} else {
				flag = true;
			}
		}
		for (int i = 0; i < s.length() / 2 && !flag; i++) {
			if (s.charAt(i) == s.charAt(last)) {
				found = false;
				for (int j = 0; j < set.length && !found; j++) {
					if (s.charAt(i) == set[j]) {
						found = true;
					}
				}
				if (!found) {
					flag = true;
				}
			} else if ((s.charAt(i) == 'b' && s.charAt(last) == 'd')
					|| (s.charAt(i) == 'd' && s.charAt(last) == 'b')
					|| (s.charAt(i) == 'p' && s.charAt(last) == 'q')
					|| (s.charAt(i) == 'q' && s.charAt(last) == 'p')) {
			} else {
				flag = true;
			}
			last--;
		}
		if (flag) {
			System.out.println("NIE");
		} else {
			System.out.println("TAK");
		}
	}
}