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 ye = "AbdHIMOopqTUVvWwXxY";
string yes = "AdbHIMOoqpTUVvWwXxY";
int main() {
  string str;
  cin >> str;
  int t = str.length(), c = 1, k;
  for (int i = 0, j = t - 1; i <= t / 2; i++, j--) {
    if (ye.find(str[i], 0) == ye.npos || ye.find(str[j], 0) == ye.npos) {
      c = 0;
      break;
    }
    k = ye.find(str[i]);
    if (str[j] != yes[k]) {
      c = 0;
      break;
    }
  }
  if (c == 1)
    printf("TAK\n");
  else
    printf("NIE\n");
  return 0;
}