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> mirror = {{'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'}};
bool is_spalin(const string& S) {
  int N = S.size();
  for (int i = 0, j = N - 1; i < N; ++i, --j) {
    if (mirror.count(S[i]) == 0) return false;
    if (mirror[S[i]] != S[j]) return false;
  }
  return true;
}
int main() {
  string S;
  cin >> S;
  cout << (is_spalin(S) ? "TAK" : "NIE") << endl;
  return 0;
}