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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string t = "AHIMOoTUVvWwXxY";
  string s;
  cin >> s;
  int n = s.length();
  for (int i = 0; i < (n + 1) / 2; i++) {
    if (!(s[i] == s[n - i - 1] && t.find(s[i]) != -1 ||
          s[i] == 'b' && s[n - i - 1] == 'd' ||
          s[i] == 'd' && s[n - i - 1] == 'b' ||
          s[i] == 'p' && s[n - i - 1] == 'q' ||
          s[i] == 'q' && s[n - i - 1] == 'p')) {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
  return 0;
}