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() {
  map<char, char> spogulAtt;
  string vards;
  int len;
  bool ok = true;
  spogulAtt['A'] = 'A';
  spogulAtt['b'] = 'd';
  spogulAtt['d'] = 'b';
  spogulAtt['H'] = 'H';
  spogulAtt['I'] = 'I';
  spogulAtt['M'] = 'M';
  spogulAtt['O'] = 'O';
  spogulAtt['o'] = 'o';
  spogulAtt['p'] = 'q';
  spogulAtt['q'] = 'p';
  spogulAtt['T'] = 'T';
  spogulAtt['U'] = 'U';
  spogulAtt['V'] = 'V';
  spogulAtt['v'] = 'v';
  spogulAtt['W'] = 'W';
  spogulAtt['w'] = 'w';
  spogulAtt['X'] = 'X';
  spogulAtt['x'] = 'x';
  spogulAtt['Y'] = 'Y';
  cin >> vards;
  len = vards.length();
  for (int i = 0; i <= len / 2; i++) {
    if (spogulAtt[vards[i]] != vards[len - 1 - i]) ok = false;
  }
  if (ok)
    cout << "TAK";
  else
    cout << "NIE";
  return 0;
}