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 N = 2e5 + 100;
const int sq = 350;
const long long int mod = 95542721;
const long long int inf = 1e16;
bool mark[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  string a;
  cin >> a;
  int n = a.size();
  mark['A'] = mark['H'] = mark['I'] = mark['M'] = mark['O'] = mark['o'] =
      mark['T'] = mark['U'] = true;
  mark['V'] = mark['v'] = mark['W'] = mark['w'] = mark['X'] = mark['x'] =
      mark['Y'] = true;
  for (int i = 0; i < n; i++) {
    if ((a[i] == a[n - i - 1] && mark[a[i]]) ||
        (a[i] == 'p' && a[n - i - 1] == 'q') ||
        (a[i] == 'b' && a[n - i - 1] == 'd'))
      continue;
    if ((a[i] == 'q' && a[n - i - 1] == 'p') ||
        (a[i] == 'd' && a[n - i - 1] == 'b'))
      continue;
    cout << "NIE" << endl;
    return 0;
  }
  cout << "TAK" << endl;
  return 0;
}