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;
const double PI = acos(-1.0);
const double EPS = 1e-9;
void task();
int main() {
  srand(time(0));
  task();
  return 0;
}
string s;
int i;
void task() {
  cin >> s;
  int sz = (int)s.size();
  set<char> sim = {'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U',
                   'V', 'v', 'W', 'w', 'X', 'x', 'Y'};
  for (i = 0; i < sz / 2 + 1; ++i)
    if ((s[i] == s[sz - i - 1] && sim.find(s[i]) != sim.end()) ||
        (s[i] == 'b' && s[sz - i - 1] == 'd') ||
        (s[i] == 'd' && s[sz - i - 1] == 'b') ||
        (s[i] == 'p' && s[sz - i - 1] == 'q') ||
        (s[i] == 'q' && s[sz - i - 1] == 'p'))
      continue;
    else {
      printf("NIE");
      return;
    }
  printf("TAK");
}