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 s0 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
string s1 = "A      HI   M O    TUVWXY  d b          oqp    vwx  ";
bool isMirroed(string str) {
  for (int i = 0, len = str.length(); i <= len / 2; i++) {
    if (s1[s0.find(str[i])] != str[len - i - 1]) return 0;
  }
  return 1;
}
int main(int argc, char const *argv[]) {
  string s;
  while (cin >> s) {
    if (isMirroed(s))
      cout << "TAK" << endl;
    else
      cout << "NIE" << endl;
  }
}