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>
char a[] = "AbdHIMOopqTUVvWwXxY", b[] = "AdbHIMOoqpTUVvWwXxY", r[300];
using namespace std;
int main() {
  int i;
  for (i = 0; i < 19; i++) r[a[i]] = b[i];
  string s;
  cin >> s;
  for (i = 0; i < s.size(); i++)
    if (r[s[i]] != s[s.size() - i - 1]) return puts("NIE"), 0;
  puts("TAK");
}