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;
map<char, char> mp;
char s[1005];
int main() {
  mp['A'] = 'A';
  mp['b'] = 'd';
  mp['d'] = 'b';
  mp['H'] = 'H';
  mp['I'] = 'I';
  mp['M'] = 'M';
  mp['O'] = 'O';
  mp['o'] = 'o';
  mp['p'] = 'q';
  mp['q'] = 'p';
  mp['T'] = 'T';
  mp['U'] = 'U';
  mp['V'] = 'V';
  mp['v'] = 'v';
  mp['W'] = 'W';
  mp['w'] = 'w';
  mp['X'] = 'X';
  mp['x'] = 'x';
  mp['Y'] = 'Y';
  scanf(" %s", s);
  int n = strlen(s);
  int l = 0, r = n - 1;
  for (int i = 0; i < n / 2; i++) {
    if (mp[s[l]] != s[r]) {
      printf("NIE\n");
      return 0;
    }
    l++;
    r--;
  }
  if (n % 2 != 0) {
    if (mp[s[n / 2]] != s[n / 2]) {
      printf("NIE\n");
      return 0;
    }
  }
  printf("TAK\n");
  return 0;
}