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;
char d[20] = {'A', 'H', 'I', 'M', 'o', 'O', 'T', 'U',
              'v', 'V', 'W', 'w', 'X', 'x', 'Y'};
char s[1005];
bool check(char c) {
  bool flag = false;
  for (int i = 0; i < 15; i++) {
    if (d[i] == c) {
      flag = true;
      break;
    }
  }
  return flag;
}
int main() {
  while (~scanf("%s", s)) {
    bool flag = true;
    int len = strlen(s), l = len / 2;
    if ((len & 1) && !check(s[l])) flag = false;
    for (int i = 0; i < l; i++) {
      if ((s[i] == 'b' && s[len - i - 1] == 'd') ||
          (s[i] == 'd' && s[len - i - 1] == 'b') ||
          (s[i] == 'p' && s[len - i - 1] == 'q') ||
          (s[i] == 'q' && s[len - i - 1] == 'p'))
        continue;
      if (check(s[i]) && check(s[len - i - 1]) && s[i] == s[len - i - 1])
        continue;
      flag = false;
      break;
    }
    if (flag)
      printf("TAK\n");
    else
      printf("NIE\n");
  }
  return 0;
}