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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, t, i, x, y;
  string s;
  cin >> s;
  map<char, bool> mp;
  mp['H'] = true;
  mp['I'] = true;
  mp['M'] = true;
  mp['T'] = true;
  mp['O'] = true;
  mp['U'] = true;
  mp['V'] = true;
  mp['W'] = true;
  mp['A'] = true;
  mp['Y'] = true;
  mp['o'] = true;
  mp['v'] = true;
  mp['x'] = true;
  mp['w'] = true;
  mp['X'] = true;
  n = s.length();
  for (i = 0; i < (n / 2); i++) {
    if ((s[i] == 'b' && s[n - i - 1] == 'd') ||
        (s[i] == 'p' && s[n - i - 1] == 'q') ||
        (s[i] == 'd' && s[n - i - 1] == 'b') ||
        (s[i] == 'q' && s[n - i - 1] == 'p') ||
        (mp[s[i]] && s[i] == s[n - i - 1])) {
      continue;
    } else {
      cout << "NIE";
      return 0;
    }
  }
  if (((n & 1) && mp[s[n / 2]]) || (n % 2 == 0)) {
    cout << "TAK";
  } else {
    cout << "NIE";
  }
}