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;
bool leap(int x) { return (x % 400 == 0 || (x % 4 == 0 && x % 100)); }
int main() {
  map<char, char> mp;
  {
    mp['A'] = 'A';
    mp['b'] = 'd', mp['d'] = 'b';
    mp['H'] = 'H';
    mp['I'] = 'I';
    mp['M'] = 'M';
    mp['o'] = 'o';
    mp['O'] = 'O';
    mp['q'] = 'p', mp['p'] = 'q';
    mp['T'] = 'T';
    mp['V'] = 'V';
    mp['v'] = 'v';
    mp['w'] = 'w';
    mp['W'] = 'W';
    mp['X'] = 'X';
    mp['x'] = 'x';
    mp['Y'] = 'Y';
    mp['U'] = 'U';
  }
  string s;
  cin >> s;
  for (int i = 0, j = s.size() - 1; i <= j; ++i, --j)
    if (mp[s[i]] != s[j]) return cout << "NIE", 0;
  cout << "TAK";
  return 0;
}