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 B691 {
	static boolean isIn(char c, char[] a) {
		for (int i = 0; i < a.length; i++) {
			if (a[i] == c)
				return true;
		}
		return false;
	}

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

}