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)
    mir1="ovwxAHIMOTUVWXY"
    mir2="bdpq"
    mir=mir1+mir2
    for i in range(n):
        if(mir.count(s[i])==1):
            bit=1
        else:
            bit=0
            break
    if(n==1 and bit==1):
        if(mir1.count(s)==1):
            print("TAK")
        else:
            print("NIE")
    elif((n%2!=0) and bit==1):
        p=n//2
        sm=s[p]
        sf=s[0:p]
        sl=s[p+1: ]
        sl=sl.replace('b','#')
        sl=sl.replace('d','b')
        sl=sl.replace('#','d')
        sl=sl.replace('p','#')
        sl=sl.replace('q','p')
        sl=sl.replace('#','q')
        sf=sf[ : :-1]
        if(sf==sl and mir1.count(sm)==1):
            print("TAK")
        else:
            print("NIE")
    elif((n%2==0) and bit==1):
        p=n//2
        
        sf=s[0:p]
        sl=s[p: ]
        sl=sl.replace('b','#')
        sl=sl.replace('d','b')
        sl=sl.replace('#','d')
        sl=sl.replace('p','#')
        sl=sl.replace('q','p')
        sl=sl.replace('#','q')
        sf=sf[ : :-1]
        
        if(sf==sl):
            print("TAK")
        else:
            print("NIE")
    else:
        print("NIE")
    
        
        
main()