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:
public class P691B {
  public static void main(String[] args) {
    java.util.Scanner in = new java.util. Scanner(System.in);
    String[] s = in.nextLine().split("");
    int start = 0;
    int end = s.length - 1;
    boolean a = true;
    String sc1 = "pbqd";
    String sc2 = "qdpb";
    String sym = "AHIMOTUVWXYovwx";
    while (start < end && a != false) {
      if(sym.contains(s[start])) a = s[start].equals(s[end]);
      else if(sc1.contains(s[start])) {int i = sc1.indexOf(s[start]); a = s[end].equals(sc2.substring(i,i+1));}
      else a = false;
      start++;
      end--;
    }
    if(s.length == 1)
      System.out.println(sym.contains(s[0]) ? "TAK" : "NIE");
    else {
      if(s.length % 2 != 0)
        System.out.println(a && sym.contains(s[start]) ? "TAK" : "NIE");
      else
        System.out.println(a ? "TAK" : "NIE");
    }
  }
}