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