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 tem1[] = "AbdHIMOopqTUVvWwXxY";
char tem2[] = "AdbHIMOoqpTUVvWwXxY";
int main() {
  char s[1010];
  scanf("%s", &s);
  int len = strlen(s);
  int i = 0, j = len - 1;
  bool flag = 1;
  while (i <= j) {
    int k;
    for (k = 0; k < strlen(tem1); k++) {
      if (tem1[k] == s[i]) break;
    }
    if (k == strlen(tem1)) {
      flag = 0;
      break;
    }
    if (tem1[k] == s[i] && tem2[k] == s[j]) {
      i++;
      j--;
      continue;
    } else {
      flag = 0;
      break;
    }
  }
  if (flag)
    printf("TAK\n");
  else
    printf("NIE\n");
  return 0;
}