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;
bool is_reflect(char c) {
  switch (c) {
    case 'A':
    case 'H':
    case 'I':
    case 'M':
    case 'O':
    case 'o':
    case 'T':
    case 'U':
    case 'V':
    case 'v':
    case 'W':
    case 'w':
    case 'X':
    case 'x':
    case 'Y':
      return true;
  }
  return false;
}
int main() {
  string s;
  cin >> s;
  long long int n = s.length();
  bool yes = true;
  for (long long int i = 0; i < n; i++) {
    bool bd = (s[i] == 'b') && (s[n - i - 1] == 'd') ||
              (s[i] == 'd') && (s[n - i - 1] == 'b');
    bool pq = (s[i] == 'p') && (s[n - i - 1] == 'q') ||
              (s[i] == 'q') && (s[n - i - 1] == 'p');
    bool reflect = (s[i] == s[n - i - 1]) && (is_reflect(s[i]));
    if (!bd && !pq && !reflect) {
      yes = false;
      break;
    }
  }
  if (yes)
    cout << "TAK" << endl;
  else
    cout << "NIE" << endl;
}