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:
a=input()
i,j=0,len(a)-1
while j>=i and [a[i],a[j]] in [['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']]:
    i+=1
    j-=1
print('TAK' if j<i else 'NIE')