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;
map<char, int> m;
int main() {
  string s;
  cin >> s;
  m['A']++;
  m['b']++;
  m['d']++;
  m['A']++;
  m['p']++;
  m['q']++;
  m['H']++;
  m['I']++;
  m['M']++;
  m['O']++;
  m['o']++;
  m['T']++;
  m['U']++;
  m['V']++;
  m['v']++;
  m['W']++;
  m['w']++;
  m['X']++;
  m['x']++;
  m['Y']++;
  for (int i = 0; i < s.size(); i++) {
    if (m[s[i]] == 0) {
      cout << "NIE";
      return 0;
    }
  }
  int t = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] != s[s.size() - 1 - i]) {
      if ((s[i] == 'q' && s[s.size() - 1 - i] == 'p') ||
          (s[i] == 'p' && s[s.size() - 1 - i] == 'q') ||
          (s[i] == 'b' && s[s.size() - 1 - i] == 'd') ||
          (s[i] == 'd' && s[s.size() - 1 - i] == 'b'))
        continue;
      else
        t = 1;
    }
    if (s[i] == s[s.size() - 1 - i]) {
      if (s[i] == 'b' || s[i] == 'd' || s[i] == 'q' || s[i] == 'p') t = 1;
    }
  }
  if (t)
    cout << "NIE";
  else
    cout << "TAK";
  return 0;
}