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 MAX = 100005;
int main() {
  ios_base::sync_with_stdio(false);
  map<char, char> xd;
  xd['A'] = 'A';
  xd['b'] = 'd';
  xd['d'] = 'b';
  xd['H'] = 'H';
  xd['I'] = 'I';
  xd['M'] = 'M';
  xd['O'] = 'O';
  xd['o'] = 'o';
  xd['p'] = 'q';
  xd['q'] = 'p';
  xd['T'] = 'T';
  xd['U'] = 'U';
  xd['V'] = 'V';
  xd['v'] = 'v';
  xd['W'] = 'W';
  xd['w'] = 'w';
  xd['X'] = 'X';
  xd['x'] = 'x';
  xd['Y'] = 'Y';
  string s;
  cin >> s;
  for (int i = 0, j = s.size() - 1; i <= j; ++i, --j)
    if (xd[s[i]] != s[j]) {
      cout << "NIE";
      return 0;
    }
  cout << "TAK";
}