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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = 1000001000;
const int mod = 1000 * 1000 * 1000 + 7;
const int mod9 = 1000 * 1000 * 1000 + 9;
const int modr = 99990001;
const double PI = 3.1415926535897932;
double sqr(double first) { return first * first; };
long double sqr(long double first) { return first * first; };
long long sqr(long long first) { return first * first; };
long long sqr(int first) { return first * 1LL * first; };
long long gcd(long long a, long long b) {
  while (b) a %= b, swap(a, b);
  return a;
}
long long bpm(long long a, long long n = -2, long long m = mod) {
  n = n >= 0 ? n : m + n;
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * a) % m;
    a = (a * a) % m;
    n >>= 1;
  }
  return r;
}
long long dist(long long x1, long long y1, long long x2, long long y2) {
  return sqr(x1 - x2) + sqr(y1 - y2);
}
std::ostream& operator<<(std::ostream& os, pair<int, int> p) {
  return os << "{" << p.first << " " << p.second << "}";
}
const int N = 1000006;
bool good(char c, char d) {
  char r = '\0';
  switch (c) {
    case 'A':
      r = 'A';
      break;
    case 'b':
      r = 'd';
      break;
    case 'd':
      r = 'b';
      break;
    case 'H':
      r = 'H';
      break;
    case 'I':
      r = 'I';
      break;
    case 'M':
      r = 'M';
      break;
    case 'O':
      r = 'O';
      break;
    case 'o':
      r = 'o';
      break;
    case 'p':
      r = 'q';
      break;
    case 'q':
      r = 'p';
      break;
    case 'T':
      r = 'T';
      break;
    case 'U':
      r = 'U';
      break;
    case 'V':
      r = 'V';
      break;
    case 'v':
      r = 'v';
      break;
    case 'W':
      r = 'W';
      break;
    case 'w':
      r = 'w';
      break;
    case 'X':
      r = 'X';
      break;
    case 'x':
      r = 'x';
      break;
    case 'Y':
      r = 'Y';
      break;
  }
  return r == d;
}
int main() {
  string s;
  cin >> s;
  bool yes = true;
  for (int i = 0; i < s.length(); i++) {
    if (!good(s[i], s[s.length() - 1 - i])) yes = false;
  }
  puts(yes ? "TAK" : "NIE");
  return 0;
}