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 m = "AHIMOKVTUWXYovwx", n;
int i, j, k;
int main() {
  cin >> n;
  for (i = 0, j = n.length() - 1; j >= i; i++, j--) {
    for (k = 0; k < m.length(); k++)
      if (n[i] == m[k] && n[j] == m[k]) break;
    if ((k < m.length()) || (n[i] == 'b' && n[j] == 'd') ||
        (n[i] == 'd' && n[j] == 'b') || (n[i] == 'q' && n[j] == 'p') ||
        (n[i] == 'p' && n[j] == 'q'))
      continue;
    break;
  }
  if (j < i)
    cout << "TAK" << endl;
  else
    cout << "NIE" << endl;
}