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 s = "AbdHIMOopqTUVvWwXxY";
  string n;
  cin >> n;
  for (int i = 0, j = n.length() - 1; i <= j; i++, j--) {
    if (s.find(n[i]) < s.length() && s.find(n[j]) < s.length()) {
      if (n[i] == n[j] && n[i] != 'b' && n[i] != 'd' && n[i] != 'p' &&
          n[i] != 'q') {
        continue;
      } else if ((n[i] == 'b' && n[j] == 'd') || (n[i] == 'd' && n[j] == 'b') ||
                 (n[i] == 'p' && n[j] == 'q') || (n[i] == 'q' && n[j] == 'p'))
        continue;
      else {
        cout << "NIE";
        return 0;
      }
    } else {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
  return 0;
}