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 temp[305];
int main() {
  char leter[] = "AbdHIMOopqTUVvWwXxY";
  char mirror[] = "AdbHIMOoqpTUVvWwXxY";
  string s;
  cin >> s;
  int len = s.size();
  for (int i = 0; i < 19; i++) temp[leter[i]] = mirror[i];
  bool flag = true;
  for (int i = 0, j = len - 1; i <= j; i++, j--) {
    if (s[i] != temp[s[j]]) {
      flag = false;
      break;
    }
  }
  if (flag)
    printf("TAK\n");
  else
    printf("NIE\n");
  return 0;
}