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;
string s;
long long f(char x, char y) {
  return (
      x == 'A' && y == 'A' || x == 'H' && y == 'H' || x == 'I' && y == 'I' ||
      x == 'M' && y == 'M' || x == 'O' && y == 'O' || x == 'T' && y == 'T' ||
      x == 'U' && y == 'U' || x == 'V' && y == 'V' || x == 'W' && y == 'W' ||
      x == 'X' && y == 'X' || x == 'Y' && y == 'Y' || x == 'b' && y == 'd' ||
      x == 'd' && y == 'b' || x == 'o' && y == 'o' || x == 'p' && y == 'q' ||
      x == 'q' && y == 'p' || x == 'v' && y == 'v' || x == 'w' && y == 'w' ||
      x == 'x' && y == 'x');
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  long long n = s.size();
  for (long long i = 0; i < n; i++) {
    if (f(s[i], s[n - i - 1]) == 0) {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
}