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']
word = raw_input()
ans = True

if len(word) == 1 and word not in sym:
    ans = False
    print "NIE"
else:    
    for i in range(0, len(word)/2+1):
        if word[i] not in allowed:
            ans = False
            break
        if word[i] in flip:
            if word[i] == 'b' and word[len(word)-i-1] != 'd':
                ans = False
                break
            if word[i] == 'd' and word[len(word)-i-1] != 'b':
                ans = False
                break
            if word[i] == 'p' and word[len(word)-i-1] != 'q':
                ans = False
                break
            if word[i] == 'q' and word[len(word)-i-1] != 'p':
                ans = False
                break
        if word[i] in sym and word[len(word)-i-1] != word[i]:
                ans = False
                break
    print "TAK" if ans else "NIE"