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 PYTHON3):

Solution:
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy


ORDA = 97
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
def divs(n, start=1):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True
def convn(number, base):
    newnumber = 0
    while number > 0:
        newnumber += number % base
        number //= base
    return newnumber
def cdiv(n, k): return n // k + (n % k != 0)


d = {'A': 'A', 'b': 'd', 'd': 'b', 'H': 'H', 'I': 'I', 'M': 'M', 'O': 'O', 'o': 'o', 'p': 'q', 'q': 'p', 'T': 'T',
     'U': 'U', 'V': 'V', 'v': 'v', 'W': 'W', 'w': 'w', 'X': 'X', 'x': 'x', 'Y': 'Y'}
s = input()
ls = len(s)
if ls & 1:
    for i in range(len(s) // 2):
        if s[i] not in d:
            print('NIE')
            exit()
        else:
            if d[s[i]] != s[-i - 1]:
                print('NIE')
                exit()
    else:
        if s[ls // 2] not in d or s[ls // 2] != d[s[ls // 2]]:
            print('NIE')
            exit()
    print('TAK')
else:
    for i in range(len(s) // 2):
        if s[i] not in d:
            print('NIE')
            exit()
        else:
            if d[s[i]] != s[-i - 1]:
                print('NIE')
                exit()
    print('TAK')