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;
map<char, char> sym = {{'A', 'A'}, {'b', 'd'}, {'d', 'b'}, {'H', 'H'},
                       {'I', 'I'}, {'M', 'M'}, {'O', 'O'}, {'o', 'o'},
                       {'p', 'q'}, {'q', 'p'}, {'T', 'T'}, {'U', 'U'},
                       {'V', 'V'}, {'v', 'v'}, {'W', 'W'}, {'w', 'w'},
                       {'X', 'X'}, {'x', 'x'}, {'Y', 'Y'}};
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  string s;
  cin >> s;
  int len = s.size() - 1;
  for (int i = (len + 1) / 2; i >= 0; i--)
    if (sym[s[i]] != s[len - i]) {
      cout << "NIE";
      return 0;
    }
  cout << "TAK";
}