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, S = "AHIMOoTUVvWwXxY";
int main() {
  cin >> s;
  int n = s.length(), m = n;
  n--;
  if (n == 0) {
    if (S.find(s[0]) < S.length()) {
      cout << "TAK\n";
    } else
      cout << "NIE\n";
    return 0;
  }
  for (int i = 0; i < m / 2; i++) {
    if (S.find(s[i]) < S.length() && s[i] == s[n - i]) continue;
    if ((s[i] == 'p' && s[n - i] == 'q') || (s[i] == 'q' && s[n - i] == 'p') ||
        (s[i] == 'b' && s[n - i] == 'd') || (s[i] == 'd' && s[n - i] == 'b'))
      continue;
    cout << "NIE\n";
    return 0;
  }
  if (m % 2 == 0 || (m % 2 && S.find(s[m / 2]) < S.length()))
    cout << "TAK\n";
  else
    cout << "NIE";
  return 0;
}