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;
int main() {
  map<char, char> m;
  m['A'] = 'A';
  m['b'] = 'd';
  m['d'] = 'b';
  m['H'] = 'H';
  m['I'] = 'I';
  m['M'] = 'M';
  m['O'] = 'O';
  m['o'] = 'o';
  m['T'] = 'T';
  m['U'] = 'U';
  m['V'] = 'V';
  m['v'] = 'v';
  m['W'] = 'W';
  m['w'] = 'w';
  m['X'] = 'X';
  m['x'] = 'x';
  m['Y'] = 'Y';
  m['p'] = 'q';
  m['q'] = 'p';
  string s;
  cin >> s;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    if (m[s[i]] && m[s[i]] == s[n - i - 1]) {
    } else {
      cout << "NIE" << endl;
      return 0;
    }
  }
  cout << "TAK" << endl;
}