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 PYTHON):

Solution:
# cook your code here
data = {}
data['A'] = 'A'
data['b'] = 'd'
data['d'] = 'b'
data['H'] = 'H'
data['I'] = 'I'
data['M'] = 'M'
data['O'] = 'O'
data['o'] = 'o'
data['p'] = 'q'
data['q'] = 'p'
data['T'] = 'T'
data['U'] = 'U'
data['V'] = 'V'
data['v'] = 'v'
data['W'] = 'W'
data['w'] = 'w'
data['X'] = 'X'
data['x'] = 'x'
data['Y'] = 'Y'

string = raw_input()

answer = True
count = 0
length = len(string)

for i in string:
    if i not in data:
        answer = False
        break
    if data[i] != string[length-1-count]:
        answer = False
        break
    count += 1
    
if answer:
    print 'TAK'
else:
    print 'NIE'