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() {
  char s[1010];
  scanf("%s", s);
  int len = strlen(s);
  char c[19][2] = {'A', 'A', 'b', 'd', 'd', 'b', 'H', 'H', 'I', 'I',
                   'M', 'M', 'O', 'O', 'o', 'o', 'p', 'q', 'q', 'p',
                   'U', 'U', 'V', 'V', 'v', 'v', 'W', 'W', 'w', 'w',
                   'X', 'X', 'x', 'x', 'Y', 'Y', 'T', 'T'};
  int flag = 0;
  for (int i = 0; i <= len / 2; i++) {
    int flag1 = 0;
    for (int j = 0; j < 19; j++) {
      if (s[i] == c[j][0] && s[len - i - 1] == c[j][1]) {
        flag1 = 1;
        break;
      }
    }
    if (flag1 == 0) {
      printf("NIE\n");
      flag = 1;
      break;
    }
  }
  if (flag == 0) printf("TAK\n");
  return 0;
}