B. s-palindrome（模拟）

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本文解析了Codeforces上一道关于s-Palindrome的题目，介绍了如何判断一个字符串是否为s-Palindrome，即是否关于中间对称，并提供了详细的代码实现，包括特殊字符的对称性判断。

CF链接：http://codeforces.com/contest/691/problem/B

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

bod

Output

TAK

Input

ER

Output

NIE

题意：要求对称的回文字符串，我们要知道哪个字母是轴对称的：

A H I M O o T U V v W w x X Y

特殊的还有p,q和b,d注意一下。

#include<iostream>
using namespace std;
int check(char s)
{
    if(s=='A'||s=='H'||s=='I'||s=='M'||s=='O'||s=='o')
        return 1;
    if(s=='T'||s=='U'||s=='V'||s=='v'||s=='W'||s=='w')
        return 1;
    if(s=='X'||s=='x'||s=='Y')
        return 1;
    return 0;
}
int main()
{
    string s;
    cin>>s;
    int flag=1;
    int len=s.size();
    for(int i=0;i<=len/2-1;i++)
    {
        if(s[i]!=s[len-1-i])
        {
            if(s[i]=='p'&&s[len-1-i]=='q')
                continue;
            if(s[i]=='b'&&s[len-1-i]=='d')
                continue;
            if(s[i]=='q'&&s[len-1-i]=='p')
                continue;
            if(s[i]=='d'&&s[len-1-i]=='b')
                continue;
            flag=0;
            break;
        }
        if(check(s[i])==0)
        {
            flag=0;
            break;
        }
    }
    if(len%2==1)
    {
        if(check(s[len/2])==0)
            flag=0;
    }
    if(flag==1)
        cout<<"TAK"<<endl;
    else
        cout<<"NIE"<<endl;
    return 0;
}
/*
A H I   M   O o T U  V v W w x X Y
*/