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 LEN = 300;
string st;
char rev[LEN];
int main() {
  cin >> st;
  rev['A'] = 'A';
  rev['b'] = 'd';
  rev['d'] = 'b';
  rev['H'] = 'H';
  rev['I'] = 'I';
  rev['M'] = 'M';
  rev['O'] = 'O';
  rev['o'] = 'o';
  rev['p'] = 'q';
  rev['q'] = 'p';
  rev['U'] = 'U';
  rev['T'] = 'T';
  rev['V'] = 'V';
  rev['v'] = 'v';
  rev['W'] = 'W';
  rev['w'] = 'w';
  rev['X'] = 'X';
  rev['x'] = 'x';
  rev['Y'] = 'Y';
  for (int i = 0; i < (st.length() - 1) / 2 + 1; i++)
    if (st[i] != rev[st[st.length() - i - 1]]) {
      cout << "NIE";
      return 0;
    }
  cout << "TAK";
  return 0;
}