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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  string a;
  ;
  cin >> a;
  char id[] = {'A', 'H', 'I', 'M', 'O', 'T', 'U', 'V',
               'W', 'X', 'Y', 'o', 'v', 'w', 'x'};
  set<char> identical(id, id + 15);
  for (int i = 0; i <= a.size() / 2; i++) {
    if ((a[i] == a[a.size() - 1 - i] && identical.count(a[i])) ||
        (a[i] == 'b' && a[a.size() - 1 - i] == 'd') ||
        (a[i] == 'd' && a[a.size() - 1 - i] == 'b') ||
        (a[i] == 'p' && a[a.size() - 1 - i] == 'q') ||
        (a[i] == 'q' && a[a.size() - 1 - i] == 'p')) {
    } else {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
  return 0;
}