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;
string s;
long ind;
bool bol = false;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  ind = s.length() - 1;
  for (int i = 0; i < s.length() / 2; i++) {
    if (s[i] != s[ind]) {
      if ((s[i] == 'b' && s[ind] == 'd') || (s[i] == 'd' && s[ind] == 'b') ||
          (s[i] == 'p' && s[ind] == 'q') || (s[i] == 'q' && s[ind] == 'p')) {
      } else
        bol = true;
    } else {
      if (s[ind] == 'A' || s[ind] == 'g' || s[ind] == 'H' || s[ind] == 'I' ||
          s[ind] == 'M' || s[ind] == 'O' || s[ind] == 'o' || s[ind] == 'T' ||
          s[ind] == 'U' || s[ind] == 'V' || s[ind] == 'v' || s[ind] == 'W' ||
          s[ind] == 'w' || s[ind] == 'X' || s[ind] == 'x' || s[ind] == 'Y') {
      } else
        bol = true;
    }
    ind--;
  }
  if (bol)
    cout << "NIE";
  else {
    if (s.length() % 2 == 1) {
      ind = s.length() / 2;
      if (s[ind] == 'A' || s[ind] == 'g' || s[ind] == 'H' || s[ind] == 'I' ||
          s[ind] == 'M' || s[ind] == 'O' || s[ind] == 'o' || s[ind] == 'T' ||
          s[ind] == 'U' || s[ind] == 'V' || s[ind] == 'v' || s[ind] == 'W' ||
          s[ind] == 'w' || s[ind] == 'X' || s[ind] == 'x' || s[ind] == 'Y')
        cout << "TAK";
      else
        cout << "NIE";
    } else
      cout << "TAK";
  }
}