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 paired[1000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  paired['A'] = 'A';
  paired['b'] = 'd';
  paired['d'] = 'b';
  paired['X'] = 'X';
  paired['x'] = 'x';
  paired['O'] = 'O';
  paired['o'] = 'o';
  paired['H'] = 'H';
  paired['V'] = 'V';
  paired['v'] = 'v';
  paired['U'] = 'U';
  paired['I'] = 'I';
  paired['T'] = 'T';
  paired['Y'] = 'Y';
  paired['W'] = 'W';
  paired['w'] = 'w';
  paired['p'] = 'q';
  paired['q'] = 'p';
  paired['M'] = 'M';
  if (s.length() % 2 == 0) {
    for (int i = 0; i < s.length(); ++i)
      if (s[i] != paired[s[s.length() - 1 - i]]) {
        cout << "NIE";
        return 0;
      }
    cout << "TAK";
    return 0;
  } else {
    for (int i = 0; i < s.length(); ++i)
      if (s[i] != paired[s[s.length() - 1 - i]]) {
        cout << "NIE";
        return 0;
      }
    cout << "TAK";
    return 0;
  }
  return 0;
}