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;
  char a[19] = {'A', 'b', 'd', 'H', 'I', 'M', 'O', 'o', 'p', 'q',
                'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y'};
  char b[19] = {'A', 'd', 'b', 'H', 'I', 'M', 'O', 'o', 'q', 'p',
                'T', 'U', 'V', 'v', 'W', 'w', 'X', 'x', 'Y'};
  map<char, char> mirror;
  for (int i = 0; i < 19; i++) {
    mirror[a[i]] = b[i];
  }
  bool p = true;
  for (int i = 0; i <= s.length() / 2; i++) {
    if (s[s.length() - 1 - i] != mirror[s[i]]) p = false;
  }
  cout << ((p == true) ? "TAK\n" : "NIE\n");
  return 0;
}