s-palindrome （模拟）

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.

Examples

Input

oXoxoXo

Output

TAK

Input

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bod

Output

TAK

Input

ER

Output

NIE

这里要注意的是，p要和q对应，b要和d对应，其余的就必须是自己就对称的字母

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
char p[15]={'A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y'};
bool f(char x){//看看在不在p数组里 
	for(int i=0;i<15;i++){
		if(x==p[i])
		    return true;
	}
	return false;
}
int main(){
	string str;
	cin>>str;
	int len=str.length() ;
    bool flag=true;
    for(int i=0;i<=len/2;i++){
    	if(str[i]==str[len-i-1]){//两个字符相等 
            if(!f(str[i]))
                flag=false;
		}
		else{
			if((str[i]=='p'&&str[len-i-1]=='q')||(str[i]=='q'&&str[len-i-1]=='p'));
			else if((str[i]=='b'&&str[len-i-1]=='d')||(str[i]=='d'&&str[len-i-1]=='b'));
			else flag=false;
		}
	}
    if(flag)
	    cout<<"TAK"<<endl;
	else
	    cout<<"NIE" <<endl;
	return 0;
}