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;
initializer_list<pair<char, char> > pairs = {
    {'A', 'A'}, {'H', 'H'}, {'I', 'I'}, {'M', 'M'}, {'O', 'O'},
    {'T', 'T'}, {'U', 'U'}, {'V', 'V'}, {'W', 'W'}, {'X', 'X'},
    {'Y', 'Y'}, {'b', 'd'}, {'d', 'b'}, {'o', 'o'}, {'p', 'q'},
    {'q', 'p'}, {'v', 'v'}, {'w', 'w'}, {'x', 'x'},
};
int main() {
  ios_base::sync_with_stdio(0);
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    char v = '!';
    for (auto j : pairs) {
      if (j.first == s[i]) v = j.second;
    }
    if (v != s[s.size() - 1 - i]) {
      cout << ("NIE");
      return 0;
    }
  }
  cout << ("TAK");
  return 0;
}