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:
def main():
    s=input()
    n=len(s)
    i=0
    j=n-1
    ans=0
    while(i<n/2):
        if(s[i]==s[j] and (s[i]=='A' or s[i]== 'H' or s[i]=='I' or s[i]=='M' or s[i]=='O' or s[i]=='T' or s[i]=='U' or s[i]=='V' or s[i]=='W' or s[i]=='X' or s[i]=='Y'or s[i]=='o' or s[i]=='v' or s[i]=='w' or s[i]=='x')):
            i += 1
            j -= 1
            continue
        elif(s[i]!=s[j] and ((s[i]=='p' and s[j]=='q') or (s[i]=='b' and s[j]=='d') or (s[i]=='d' and s[j]=='b') or (s[i]=='q' and s[j]=='p'))):
            i += 1
            j -= 1
            continue
        else:
            ans=1
        i+=1
        j-=1
    if(ans):
        print("NIE")
    else:
        print("TAK")
if __name__ == '__main__':main()