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.Locale;
import java.util.StringTokenizer;

public class A implements Runnable {

    private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tok = new StringTokenizer("");

    private void init() throws FileNotFoundException {
        Locale.setDefault(Locale.US);
        String fileName = "";
        if (ONLINE_JUDGE && fileName.isEmpty()) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            try {
                if (fileName.isEmpty()) {
                    in = new BufferedReader(new FileReader("input.txt"));
                    out = new PrintWriter("output.txt");
                } else {
                    in = new BufferedReader(new FileReader(fileName + ".in"));
                    out = new PrintWriter(fileName + ".out");
                }
            } catch (Throwable t) {
                in = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            }
        }
    }

    String readString() {
        while (!tok.hasMoreTokens()) {
            try {
                tok = new StringTokenizer(in.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return tok.nextToken();
    }

    int readInt() {
        return Integer.parseInt(readString());
    }

    long readLong() {
        return Long.parseLong(readString());
    }

    double readDouble() {
        return Double.parseDouble(readString());
    }

    int[] readIntArray(int size) {
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = readInt();
        }
        return a;
    }

    public static void main(String[] args) {
        //new Thread(null, new _Solution(), "", 128 * (1L << 20)).start();
        new A().run();
    }

    long timeBegin, timeEnd;

    void time() {
        timeEnd = System.currentTimeMillis();
        System.err.println("Time = " + (timeEnd - timeBegin));
    }

    @Override
    public void run() {
        try {
            timeBegin = System.currentTimeMillis();
            init();
            solve();
            out.close();
            time();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    String selfGood = "AHIMOoTUVvWwXxY";

    private void solve() {
        char[] s = readString().toCharArray();
        boolean yes = true;
        for (int i = 0; i < s.length; i++) {
            if (s[i] == s[s.length - i - 1]) {
                if (!selfGood.contains("" + s[i])) {
                    yes = false;
                }
            } else {
                if (s[i] == 'p' && s[s.length - i - 1] == 'q') continue;
                if (s[i] == 'q' && s[s.length - i - 1] == 'p') continue;
                if (s[i] == 'b' && s[s.length - i - 1] == 'd') continue;
                if (s[i] == 'd' && s[s.length - i - 1] == 'b') continue;
                yes = false;
            }
        }
        out.println(yes ? "TAK" : "NIE");
    }
}