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;
const char ok[] = "AbdHIMOopqTUVvWwXxY";
const char mr[] = "AdbHIMOoqpTUVvWwXxY";
char s[1005], t[1005];
int main() {
  int tot = strlen(ok);
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; i++) {
    bool isok = 0;
    for (int j = 0; j < tot; j++)
      if (s[i] == ok[j]) {
        t[n - 1 - i] = mr[j];
        isok = 1;
      }
    if (!isok) return 0 * printf("NIE");
  }
  printf("%s", (strcmp(s, t) == 0 ? "TAK" : "NIE"));
  return 0;
}