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>
#pragma comment(linker, "/STACK:10000000000")
using namespace std;
const int maxN = 1e6, maxM = 10, alpha = 4;
const int INF = 1e9, mod = 1e9 + 9;
string center = "AHIMOoTUVvWwXxY";
int main() {
  string s;
  cin >> s;
  if (s.size() % 2) {
    if (center.find(s[s.size() / 2]) == -1) {
      cout << "NIE";
      return 0;
    }
  }
  for (int i = 0; i < s.size() / 2; i++) {
    if (s[i] == 'p' && s[s.size() - 1 - i] == 'q' ||
        s[i] == 'q' && s[s.size() - 1 - i] == 'p') {
      continue;
    }
    if (s[i] == 'b' && s[s.size() - 1 - i] == 'd' ||
        s[i] == 'd' && s[s.size() - 1 - i] == 'b') {
      continue;
    }
    if (s[i] != s[s.size() - 1 - i] || center.find(s[i]) == -1) {
      cout << "NIE";
      return 0;
    }
  }
  cout << "TAK";
  return 0;
}