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:
import sys

pairs = [
    'AA', 'HH', 'II', 'MM', 'OO', 'TT', 'UU', 'VV', 'WW', 'XX', 'YY',
    'bd', 'db', 'oo', 'pq', 'qp', 'vv', 'ww', 'xx'
]

s = input()

l, r = 0, len(s) - 1
while l <= r:
    if s[l] + s[r] not in pairs:
        print("NIE")
        sys.exit(0)
    l += 1
    r -= 1

print("TAK")