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:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Jul 13 10:33:00 2020

@author: shailesh
"""

s = input()

mirror_letters = ['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']#check i,

mirror_d = {}

for char in mirror_letters:
    mirror_d[char] = char

mirror_d['b'] = 'd'
mirror_d['d'] = 'b'
mirror_d['p'] = 'q'
mirror_d['q'] = 'p'

flag = 1
n = len(s)
for i in range((n+1)//2):
    if s[i] in mirror_d and s[n-1-i] == mirror_d[s[i]]:
        continue
    else:
        flag = 0
        break

if flag:
    print('TAK')
else:
    print('NIE')