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
#sys.stdin = open('in', 'r')
#n = int(input())
#a = [int(x) for x in input().split()]
#n,m = map(int, input().split())

s = input()

syms = {"A": "A", "b": "d", "d": "b", "H": "H", 'I':'I', 'M':'M','O':'O','o':'o', 'T':'T', 'U':'U', 'V':'V', 'v':'v', 'W':'W', 'w':'w', 'X':'X', 'x':'x', 'Y':'Y', 'p':'q','q':'p'}

if (len(s) % 2 == 1 and s[len(s)//2] in syms and s[len(s)//2] not in ['b', 'd', 'p', 'q']) or len(s) % 2 == 0:
    yes = True
    for i in range(len(s)//2):
        if s[i] in syms and syms[s[i]] == s[-i-1]:
            yes &= True
        else:
            yes = False
    print("TAK" if yes else "NIE")
else:
    print("NIE")