CodeForces - 691B s-palindrome (字符串，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.

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");
    }
}