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 c[123];
int main() {
  fill(c, c + 123, '.');
  string s;
  cin >> s;
  set<char> p = {'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U',
                 'V', 'W', 'w', 'v', 'X', 'x', 'Y'};
  c['A'] = 'A';
  c['b'] = 'd';
  c['d'] = 'b';
  c['H'] = 'H';
  c['I'] = 'I';
  c['M'] = 'M';
  c['O'] = 'O';
  c['o'] = 'o';
  c['p'] = 'q';
  c['q'] = 'p';
  c['T'] = 'T';
  c['U'] = 'U';
  c['V'] = 'V';
  c['v'] = 'v';
  c['W'] = 'W';
  c['w'] = 'w';
  c['X'] = 'X';
  c['x'] = 'x';
  c['Y'] = 'Y';
  for (int i = 0; i < (s.size() / 2); i++) {
    if (c[s[i]] != s[(int)s.size() - 1 - i]) {
      cout << "NIE";
      return 0;
    }
  }
  if (s.size() % 2 == 0) {
    cout << "TAK";
    return 0;
  }
  char x = s[(int)(s.size() / 2)];
  if (p.find(x) != p.end())
    cout << "TAK";
  else
    cout << "NIE";
}