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>
char f[255], s[1005];
int main() {
  f['A'] = 'A';
  f['H'] = 'H';
  f['I'] = 'I';
  f['M'] = 'M';
  f['O'] = 'O';
  f['T'] = 'T';
  f['U'] = 'U';
  f['V'] = 'V';
  f['W'] = 'W';
  f['X'] = 'X';
  f['Y'] = 'Y';
  f['b'] = 'd';
  f['d'] = 'b';
  f['o'] = 'o';
  f['p'] = 'q';
  f['q'] = 'p';
  f['v'] = 'v';
  f['w'] = 'w';
  f['x'] = 'x';
  scanf("%s", s);
  for (int i = 0, j = strlen(s) - 1; i <= j; ++i, --j)
    if (f[s[i]] != s[j]) {
      puts("NIE");
      return 0;
    }
  puts("TAK");
}