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;
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto it : v) out << it << ' ';
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &it : v) in >> it;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
  out << p.first << ' ' << p.second << ' ';
  return out;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  set<char> sim({'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W', 'w',
                 'X', 'x', 'Y'});
  string s;
  cin >> s;
  for (int i = 0; i < (s.size() + 1) / 2; i++) {
    if (sim.count(s[i]) && s[s.size() - i - 1] == s[i]) continue;
    if (s[i] == 'd' && s[s.size() - i - 1] == 'b') continue;
    if (s[i] == 'b' && s[s.size() - i - 1] == 'd') continue;
    if (s[i] == 'p' && s[s.size() - i - 1] == 'q') continue;
    if (s[i] == 'q' && s[s.size() - i - 1] == 'p') continue;
    cout << "NIE";
    return 0;
  }
  cout << "TAK";
  return 0;
}