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;
char sym(char a) {
  if (a == 'A') {
    return 'A';
  }
  if (a == 'b') {
    return 'd';
  }
  if (a == 'd') {
    return 'b';
  }
  if (a == 'H') {
    return 'H';
  }
  if (a == 'I') {
    return 'I';
  }
  if (a == 'M') {
    return 'M';
  }
  if (a == 'O') {
    return 'O';
  }
  if (a == 'o') {
    return 'o';
  }
  if (a == 'p') {
    return 'q';
  }
  if (a == 'q') {
    return 'p';
  }
  if (a == 'T') {
    return 'T';
  }
  if (a == 'U') {
    return 'U';
  }
  if (a == 'V') {
    return 'V';
  }
  if (a == 'v') {
    return 'v';
  }
  if (a == 'W') {
    return 'W';
  }
  if (a == 'w') {
    return 'w';
  }
  if (a == 'X') {
    return 'X';
  }
  if (a == 'x') {
    return 'x';
  }
  if (a == 'Y') {
    return 'Y';
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); ++i)
    if (s[s.size() - i - 1] != sym(s[i])) {
      cout << "NIE" << endl;
      return 0;
    }
  cout << "TAK" << endl;
}