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 int MAX = 1010;
char s[MAX];
int main() {
  set<char> simetric({'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U', 'V', 'v', 'W',
                      'w', 'X', 'x', 'Y'});
  scanf("%s", s);
  int n = strlen(s);
  if (n % 2 && !simetric.count(s[n / 2])) {
    puts("NIE");
    return 0;
  }
  for (int i = (0); i < (n / 2); ++i) {
    if (simetric.count(s[i]) && s[i] == s[n - i - 1]) continue;
    if (s[i] == 'p' && s[n - i - 1] == 'q') continue;
    if (s[i] == 'q' && s[n - i - 1] == 'p') continue;
    if (s[i] == 'b' && s[n - i - 1] == 'd') continue;
    if (s[i] == 'd' && s[n - i - 1] == 'b') continue;
    puts("NIE");
    return 0;
  }
  puts("TAK");
  return 0;
}