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;
  cin >> s;
  set<char> x;
  x.insert('A');
  x.insert('H');
  x.insert('I');
  x.insert('M');
  x.insert('O');
  x.insert('T');
  x.insert('U');
  x.insert('V');
  x.insert('W');
  x.insert('X');
  x.insert('Y');
  x.insert('o');
  x.insert('v');
  x.insert('x');
  x.insert('w');
  int i, j, n = s.length();
  i = 0, j = n - 1;
  for (; i <= j; i++, j--) {
    if (!((s[i] == s[j] && x.find(s[i]) != x.end()) ||
          (s[i] == 'b' && s[j] == 'd') || (s[i] == 'd' && s[j] == 'b') ||
          (s[i] == 'p' && s[j] == 'q') || (s[i] == 'q' && s[j] == 'p'))) {
      cout << "NIE" << endl;
      break;
    }
  }
  if (i == j) {
    if (x.find(s[i]) != x.end()) cout << "TAK" << endl;
  }
  if (i > j) cout << "TAK" << endl;
  return 0;
}