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 n, m, i, k = 200;
string s;
map<char, char> p;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s;
  for (char i = 'a'; i <= 'z'; i++) p[i] = char(k), k++;
  for (char i = 'A'; i <= 'Z'; i++) p[i] = char(k), k++;
  n = s.size();
  p['X'] = 'X';
  p['x'] = 'x';
  p['H'] = 'H';
  p['T'] = 'T';
  p['M'] = 'M';
  p['A'] = 'A';
  p['I'] = 'I';
  p['O'] = 'O';
  p['o'] = 'o';
  p['Y'] = 'Y';
  p['U'] = 'U';
  p['W'] = 'W';
  p['w'] = 'w';
  p['v'] = 'v';
  p['V'] = 'V';
  p['q'] = 'p';
  p['p'] = 'q';
  p['b'] = 'd';
  p['d'] = 'b';
  for (i = 0; i <= n / 2; i++) {
    if (p[s[i]] != s[n - i - 1] || s[i] != p[s[n - i - 1]]) {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
}