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;
const int N = 10005;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  bool b = 0;
  string sp = "AHIMOTUVWXYovwxbdpq";
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    if (sp.find(s[i]) == -1) {
      b = 1;
      break;
    } else if (s[i] == 'b' && s[(s.size() - 1) - i] != 'd') {
      b = 1;
      break;
    } else if (s[i] == 'd' && s[(s.size() - 1) - i] != 'b') {
      b = 1;
      break;
    } else if (s[i] == 'p' && s[(s.size() - 1) - i] != 'q') {
      b = 1;
      break;
    } else if (s[i] == 'q' && s[(s.size() - 1) - i] != 'p') {
      b = 1;
      break;
    } else if (s[i] != s[(s.size() - 1) - i]) {
      if ((s[i] != 'b' && s[(s.size() - 1) - i] != 'd') &&
          (s[i] != 'd' && s[(s.size() - 1) - i] != 'b') &&
          (s[i] != 'p' && s[(s.size() - 1) - i] != 'q') &&
          (s[i] != 'q' && s[(s.size() - 1) - i] != 'p')) {
        b = 1;
        break;
      }
    }
  }
  if (!b)
    cout << "TAK\n";
  else
    cout << "NIE\n";
  return 0;
}