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