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;
const int N = (int)1e5 + 113;
set<char> sim;
map<char, char> par;
string s;
int main() {
  cin >> s;
  sim.insert('A');
  sim.insert('H');
  sim.insert('I');
  sim.insert('M');
  sim.insert('O');
  sim.insert('o');
  sim.insert('T');
  sim.insert('U');
  sim.insert('V');
  sim.insert('v');
  sim.insert('W');
  sim.insert('w');
  sim.insert('X');
  sim.insert('x');
  sim.insert('Y');
  par['d'] = 'b';
  par['b'] = 'd';
  par['q'] = 'p';
  par['p'] = 'q';
  for (int i = 0; i < s.size(); i++) {
    int p = i;
    int q = s.size() - 1 - i;
    if (s[p] == s[q]) {
      if (sim.find(s[p]) != sim.end()) continue;
      cout << "NIE";
      return 0;
    }
    if (par[s[p]] == s[q]) continue;
    cout << "NIE";
    return 0;
  }
  cout << "TAK";
}