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() {
  int i, l;
  char s[1001], t[256];
  memset(t, 0, sizeof(t));
  t['A'] = 'A';
  t['b'] = 'd';
  t['d'] = 'b';
  t['H'] = 'H';
  t['I'] = 'I';
  t['M'] = 'M';
  t['o'] = 'o';
  t['O'] = 'O';
  t['p'] = 'q';
  t['q'] = 'p';
  t['T'] = 'T';
  t['U'] = 'U';
  t['V'] = 'V';
  t['v'] = 'v';
  t['W'] = 'W';
  t['w'] = 'w';
  t['X'] = 'X';
  t['x'] = 'x';
  t['Y'] = 'Y';
  scanf("%s", s);
  l = strlen(s);
  for (i = 0; i < l && t[s[i]] == s[l - 1 - i]; i++) {
  }
  printf(i < l ? "NIE\n" : "TAK\n");
  return 0;
}