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;
long long i;
string s, ss;
int main() {
  cin >> s;
  for (i = 0; i < s.size() / 2 + s.size() % 2; i++) {
    ss = s[i];
    ss += s[s.size() - i - 1];
    if (ss != "AA" && ss != "bd" && ss != "db" && ss != "oo" && ss != "OO" &&
        ss != "pq" && ss != "qp" && ss != "xx" && ss != "XX" && ss != "ww" &&
        ss != "WW" && ss != "vv" && ss != "VV" && ss != "VV" && ss != "TT" &&
        ss != "YY" && ss != "MM" && ss != "UU" && ss != "HH" && ss != "II") {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
  return 0;
}