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 MAXN = 1e5 + 10;
string s;
map<char, char> pal;
int main() {
  cin >> s;
  pal['A'] = 'A';
  pal['b'] = 'd';
  pal['d'] = 'b';
  pal['H'] = 'H';
  pal['I'] = 'I';
  pal['M'] = 'M';
  pal['O'] = 'O';
  pal['o'] = 'o';
  pal['T'] = 'T';
  pal['U'] = 'U';
  pal['V'] = 'V';
  pal['v'] = 'v';
  pal['W'] = 'W';
  pal['w'] = 'w';
  pal['X'] = 'X';
  pal['x'] = 'x';
  pal['Y'] = 'Y';
  pal['q'] = 'p';
  pal['p'] = 'q';
  for (int i = 0, _b = (s.size() / 2); i <= _b; i++)
    if (pal.find(s[i]) == pal.end() || s[s.size() - i - 1] != pal[s[i]]) {
      cout << "NIE";
      return 0;
    }
  cout << "TAK";
}