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:
#!/usr/bin/python3.4
a="AHIMOTUVWXYovwx"
b={'b':'d', 'd':'b', 'p':'q', 'q':'p'}
s=str(input())

ok = True
slen = len(s)
for i in range((slen - 1) // 2 + 1):
	if (s[i] in a and s[i] == s[slen - i - 1]) or (b.get(s[i], 0) and b[s[i]] == s[slen - 1 - i]):
		continue
	else:
		ok = False
		break;
print(['NIE', 'TAK'][ok])