s-palindrome

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D - s-palindrome

Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

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.

$\text{[math]}$ 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.

Sample Input

Input

oXoxoXo

Output

TAK

Input

bod

Output

TAK

Input

ER

Output

NIE

大概题意就是：轴对称的字符串输出TAK ，否则输出ER

#include <bits/stdc++.h>
using namespace std;
int main()
{
    string s;
    cin>>s;
    int len=s.size();
    char c[19][2]={'A','A','b','d','d','b','H','H','I','I','M','M','O','O','o',
    'o','p','q','q','p','U','U','V','V','v','v','W','W','w','w','X','X','x','x','Y','Y','T','T'};
    int flag=0;
    for(int i=0;i<=len/2;i++)
    {
        int fg=0;
        for(int j=0;j<19;j++)
        {
            if(s[i]==c[j][0]&&s[len-i-1]==c[j][1])
            {
                fg=1;
                break;
            }
        }
        if(fg==0)
        {
            puts("NIE");
            flag=1;
            break;
        }
    }
    if(flag==0) puts("TAK");
    return 0;
}

岂能尽如人意，但求无愧我心