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;
  char check[1000], s[10001];
  char a[] = "AbdHIMOopqTUVvWwXxY";
  char b[] = "AdbHIMOoqpTUVvWwXxY";
  for (i = 0; a[i]; i++) check[a[i]] = b[i];
  scanf("%s", s);
  for (i = 0; i < strlen(s); i++) {
    if (check[s[i]] != s[strlen(s) - i - 1]) {
      printf("NIE\n");
      return 0;
    }
  }
  printf("TAK\n");
  return 0;
}