CodeForces

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.

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.

Example

Input

oXoxoXo

Output

TAK

Input

bod

Output

TAK

Input

ER

Output

NIE

题意：判断一个字符串是否是对称的，是的话输出TAK，否则输出NIE。

首先要找出对称的字母，注意m、n两个字母不对称！

还要注意，b和d对称，p和q对称！

另外注意字符串对称的写法！

代码如下：

#include<bits/stdc++.h>using namespace std;int main(){    char *s="AHIMOoTUVvWwXxY";    char a[1010];    bool flag=true;    scanf("%s",a);    for(int i=0;i<=strlen(a)/2;i++)    {        if(a[i]!=a[strlen(a)-i-1])        {            if((a[i]=='b'&&a[strlen(a)-i-1]=='d')||(a[i]=='d'&&a[strlen(a)-i-1]=='b'));            else if((a[i]=='p'&&a[strlen(a)-i-1]=='q')||(a[i]=='q'&&a[strlen(a)-i-1]=='p'));            else flag=false;        }       if((a[i]=='b'&&a[strlen(a)-i-1]=='d')||(a[i]=='d'&&a[strlen(a)-i-1]=='b'));            else if((a[i]=='p'&&a[strlen(a)-i-1]=='q')||(a[i]=='q'&&a[strlen(a)-i-1]=='p'));//这样写是为了能判断到中间的字母是否对称，但是没有中间字母的时候会错开        else if(strchr(s,a[i])==NULL)        {            flag=false;        }    }    if(flag==true)    {        printf("TAK\n");    }    else    {        printf("NIE\n");    }}