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>
int main() {
  std::string s;
  std::cin >> s;
  int start = 0;
  int end = s.length() - 1;
  while (start <= end) {
    switch (s[start]) {
      case 'A':
      case 'H':
      case 'I':
      case 'M':
      case 'O':
      case 'o':
      case 'T':
      case 'U':
      case 'V':
      case 'v':
      case 'W':
      case 'w':
      case 'X':
      case 'x':
      case 'Y':
        if (s[end] != s[start]) {
          std::cout << "NIE" << std::endl;
          return 0;
        }
        end--;
        start++;
        break;
      case 'b':
        if (s[end] != 'd') {
          std::cout << "NIE" << std::endl;
          return 0;
        }
        end--;
        start++;
        break;
      case 'd':
        if (s[end] != 'b') {
          std::cout << "NIE" << std::endl;
          return 0;
        }
        end--;
        start++;
        break;
      case 'p':
        if (s[end] != 'q') {
          std::cout << "NIE" << std::endl;
          return 0;
        }
        end--;
        start++;
        break;
      case 'q':
        if (s[end] != 'p') {
          std::cout << "NIE" << std::endl;
          return 0;
        }
        end--;
        start++;
        break;
      default:
        std::cout << "NIE" << std::endl;
        return 0;
    }
  }
  std::cout << "TAK" << std::endl;
  return 0;
}