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;
bool check(char a, char b) {
  if (a > b) swap(a, b);
  if (a == 'A' || a == 'H' || a == 'I' || a == 'M' || a == 'O' || a == 'o' ||
      a == 'T' || a == 'U' || a == 'v' || a == 'V' || a == 'w' || a == 'W' ||
      a == 'x' || a == 'X' || a == 'Y')
    return b == a;
  else if (a == 'b')
    return b == 'd';
  else if (a == 'p')
    return b == 'q';
  return false;
}
int main() {
  string s;
  cin >> s;
  int l = (int)s.size();
  bool flag = true;
  for (int i = 0; i < l / 2; ++i) {
    if (!check(s[i], s[l - 1 - i])) {
      flag = false;
      break;
    }
  }
  if (l & 1) flag = flag & check(s[l / 2], s[l / 2]);
  puts(flag ? "TAK" : "NIE");
  return 0;
}