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;
void init() {}
const int N = 63e5 + 34;
const long long INF = 2e9 + 34;
int MD = 30;
string sym = "AHIMOTUVWXYovwx";
void fail() {
  cout << "NIE";
  exit(0);
}
int main() {
  init();
  string s;
  cin >> s;
  int n = s.length();
  if (n % 2 && sym.find(s[n / 2]) > 3000) fail();
  for (int i = 0; i < n / 2; i++) {
    if (sym.find(s[i]) < 3000 && sym.find(s[n - i - 1]) < 300 &&
        s[i] == s[n - i - 1])
      continue;
    if (s[i] == 'b' && s[n - i - 1] == 'd') continue;
    if (s[i] == 'd' && s[n - i - 1] == 'b') continue;
    if (s[i] == 'p' && s[n - i - 1] == 'q') continue;
    if (s[i] == 'q' && s[n - i - 1] == 'p') continue;
    fail();
  }
  cout << "TAK";
  return 0;
}