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;
char cc[] = {'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U',
             'V', 'v', 'W', 'w', 'X', 'x', 'Y', '\0'};
bool jud(char x) {
  int len = strlen(cc);
  for (int i = 0; i < len; i++) {
    if (x == cc[i]) return true;
  }
  return false;
}
bool jud2(char x, char y) {
  if (x > y) swap(x, y);
  if (x == 'b' && y == 'd' || x == 'p' && y == 'q') {
    return true;
  }
  return false;
}
int main() {
  string s;
  cin >> s;
  int len = s.length();
  for (int i = 0; i <= len / 2; i++) {
    if (jud(s[i]) && s[i] == s[len - 1 - i] || jud2(s[i], s[len - 1 - i]))
      ;
    else {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
  return 0;
}