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:

allowed = ['A','b','d','H','I','M','O','o','p','q','T','U','V','v','W','w','X','x','Y']

sym = ['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']
flip = ['b','d','p','q']

s = raw_input()

ans = True

for i in range(0,int(len(s)/2)+1) :
    if s[i] not in allowed :
        ans = False
        break
    if i == len(s)-i-1 and s[i] not in sym :
        ans = False
        break
    if s[i] in flip :
        if s[i] == 'b' and s[len(s)-i-1] != 'd' :
            ans = False
            break
        elif s[i] == 'd' and s[len(s)-i-1] != 'b' :
            ans = False
            break
        elif s[i] == 'p' and s[len(s)-i-1] != 'q' :
            ans = False
            break
        elif s[i] == 'q' and s[len(s)-i-1] != 'p' :
            ans = False
            break
    if s[i] in sym and s[len(s)-i-1] != s[i] :
        ans = False
        break
        

if ans : print "TAK"
else : print "NIE"