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:
# Joshua Li
# Educational Codeforces Round 14 Problem B

# つわもののみいきのこます。
# わたしはさいぜんをつくすことをやくそくします。

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

s, flag = input(), True

for i in range(int(len(s)/2)+1):
    if s[i] not in selfsym:
        if s[i] in opsym.keys():
            if opsym[s[i]] != s[len(s)-i-1]:
                flag = False
        else:
            flag = False
    else:
        if s[i] != s[len(s)-i-1]:
            flag = False

if len(s) % 2 == 1:
    if s[int(len(s)/2)] not in selfsym:
        flag = False

print("TAK" if flag else "NIE")