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;
int main() {
  char A[] = {'A', 'H', 'I', 'M', 'O', 'o', 'T', 'U',
              'V', 'v', 'W', 'w', 'X', 'x', 'Y'};
  char B[] = {'b', 'p'};
  char C[] = {'d', 'q'};
  char S[1002];
  int n, i, j, k, flag = 0;
  int flag1 = 1;
  cin >> S;
  n = strlen(S);
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j <= 14; j++) {
      if (S[i] == A[j]) {
        if (S[n - 1 - i] == A[j]) flag = 1;
        break;
      } else
        flag = 0;
    }
    if (flag == 0) {
      for (k = 0; k <= 1; k++) {
        if (S[i] == B[k] && S[n - 1 - i] == C[k]) {
          flag = 1;
          break;
        }
        if (S[i] == C[k] && S[n - 1 - i] == B[k]) {
          flag = 1;
          break;
        }
      }
    }
    if (flag == 0) {
      flag1 = 0;
      break;
    }
  }
  if (n % 2 == 1 && flag1 == 1) {
    for (j = 0; j <= 14; j++) {
      if (A[j] == S[n / 2]) {
        flag1 = 1;
        break;
      } else
        flag1 = 0;
    }
  }
  if (flag1 == 0)
    cout << "NIE" << endl;
  else
    cout << "TAK" << endl;
  return 0;
}