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