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:
m=['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']
n={'p': 'q', 'q': 'p', 'b': 'd', 'd': 'b'}
a=True
x=input()
for i in range(len(x)//2):
    if x[i] in m:
        if x[-1-i]!=x[i]:
            a=False
            break
    elif x[i] in ['b','d','q','p']:
        if n[x[i]]!=x[-1-i]:
            a=False
            break
    else:
        a=False
        break
if len(x)%2==1 and x[len(x)//2] not in m:
    a=False
if a:
    print('TAK')
else:
    print('NIE')