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 ans = 0;
int main() {
  string symmetric = "AHIMOoTUVvWwXxY";
  string s;
  cin >> s;
  if (s.size() == 1 && symmetric.find(s) != -1) {
    cout << "TAK";
    return 0;
  }
  for (int i = 0, j = s.size() - 1; i < s.size() / 2; i++, j--) {
    if (s[i] == s[j] && symmetric.find(s[i]) != -1 ||
        (s[i] == 'p' && s[j] == 'q') || (s[i] == 'q' && s[j] == 'p') ||
        (s[i] == 'd' && s[j] == 'b') || (s[i] == 'b' && s[j] == 'd'))
      ans++;
  }
  if (ans == s.size() / 2 && s.size() % 2 == 0)
    cout << "TAK";
  else if (ans == s.size() / 2 && s.size() % 2 != 0 &&
           symmetric.find(s[(s.size() / 2)]) != -1)
    cout << "TAK";
  else
    cout << "NIE";
  return 0;
}