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;
const int INF = 1 << 28;
int main() {
  string a = "AHIMOoTUVvWwXxY";
  string in;
  cin >> in;
  bool flg = true;
  for (int i = 0; i < (int)in.size(); i++) {
    if (in[i] == 'b' && in[in.size() - i - 1] == 'd') {
    } else if (in[i] == 'd' && in[in.size() - i - 1] == 'b') {
    } else if (in[i] == 'p' && in[in.size() - i - 1] == 'q') {
    } else if (in[i] == 'q' && in[in.size() - i - 1] == 'p') {
    } else if (a.find(in[i]) != -1 && in[in.size() - i - 1] == in[i]) {
    } else
      flg = false;
  }
  if (flg)
    cout << "TAK" << endl;
  else
    cout << "NIE" << endl;
  return 0;
}