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);
  set<char> isMirrorLetter{'p', 'q', 'd', 'b'};
  map<char, char> mirrorLetter = {
      {'p', 'q'}, {'q', 'p'}, {'d', 'b'}, {'b', 'd'}};
  set<char> isGoodLetter{'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U',
                         'V', 'v', 'W', 'w', 'X', 'x', 'Y'};
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (isGoodLetter.find(s[i]) != isGoodLetter.end()) {
      if (s[i] != s[s.size() - i - 1]) {
        printf("NIE\n");
        return 0;
      }
    } else if (isMirrorLetter.find(s[i]) != isMirrorLetter.end()) {
      if (s[s.size() - i - 1] != mirrorLetter[s[i]]) {
        printf("NIE\n");
        return 0;
      }
    } else {
      printf("NIE\n");
      return 0;
    }
  }
  printf("TAK\n");
}