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;
template <class T, class U>
inline void chmin(T &t, U f) {
  if (t > f) t = f;
}
template <class T, class U>
inline void chmax(T &t, U f) {
  if (t < f) t = f;
}
string lis = "aBCcDEeFfGghiJjKkLlmNnPQRrSstuyZz";
signed main() {
  string s;
  cin >> s;
  bool ok = true;
  for (long long i = 0; i < s.size(); i++) {
    char a = s[i];
    char b = s[s.size() - 1 - i];
    if (find((lis).begin(), (lis).end(), a) != lis.end()) ok = false;
    if (a == 'b')
      ok &= b == 'd';
    else if (a == 'd')
      ok &= b == 'b';
    else if (a == 'p')
      ok &= b == 'q';
    else if (a == 'q')
      ok &= b == 'p';
    else
      ok &= a == b;
  }
  if (ok)
    cout << "TAK" << endl;
  else
    cout << "NIE" << endl;
  return 0;
}