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:
string = input()

D = {"A":["A"],"b":["d"],"d":["b"],"H":["H"],"I":["I"],"M":["M"],"O":["O"],"o":["o"],"p":["q"],"q":["p"],"T":["T"],"U":["U"],"V":["V"],"v":["v"],"W":["W"],"w":["w"],"X":["X"],"x":["x"],"Y":["Y"]}

flag = 0
t = 0

if len(string) % 2 == 1:
	t = len(string)//2 + 1;
else:
	t = len(string)//2

for i in range(t):
	#print (i, string[i])
	if string[i] in D:

		if D[string[i]][0] == string[len(string)-1-i]:
			continue;
		else:
			#print (i, string[i])
			flag = 1;
			break;
	else:
		flag = 1;
		break;

if flag == 1:
	print("NIE")
else:
	print("TAK")