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 MAXN = 100 * 100 * 10 * 3 + 10;
map<char, char> mp;
int main() {
  mp['A'] = 'A', mp['H'] = 'H', mp['I'] = 'I', mp['M'] = 'M', mp['O'] = 'O',
  mp['T'] = 'T', mp['U'] = 'U', mp['V'] = 'V', mp['W'] = 'W', mp['X'] = 'X',
  mp['Y'] = 'Y';
  mp['b'] = 'd', mp['d'] = 'b', mp['o'] = 'o', mp['v'] = 'v', mp['w'] = 'w',
  mp['x'] = 'x', mp['p'] = 'q', mp['q'] = 'p';
  string s;
  cin >> s;
  for (int i = 0; i * 2 < s.size(); i++)
    if (mp[s[i]] != s[s.size() - i - 1]) return cout << "NIE", 0;
  cout << "TAK" << endl;
  return 0;
}