CF286-A

A. Mr. Kitayuta's Gift

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mr. Kitayuta has kindly given you a string s consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter into s to make it a palindrome. A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not.

You can choose any lowercase English letter, and insert it to any position of s, possibly to the beginning or the end of s. You have to insert a letter even if the given string is already a palindrome.

If it is possible to insert one lowercase English letter into s so that the resulting string will be a palindrome, print the string after the insertion. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one palindrome that can be obtained, you are allowed to print any of them.

Input

The only line of the input contains a string s (1 ≤ |s| ≤ 10). Each character in s is a lowercase English letter.

Output

If it is possible to turn s into a palindrome by inserting one lowercase English letter, print the resulting string in a single line. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one solution, any of them will be accepted.

Sample test(s)

input

revive

output

reviver

input

ee

output

eye

input

kitayuta

output

NA

Note

For the first sample, insert 'r' to the end of "revive" to obtain a palindrome "reviver".

For the second sample, there is more than one solution. For example, "eve" will also be accepted.

For the third sample, it is not possible to turn "kitayuta" into a palindrome by just inserting one letter.

给出一个字符串,问是否能向其中插入一个字符使得新的字符串是个回文串

由于给出的字符串长度很短,最大只有10.故可以暴力求解

要使得插入字符后新字符串成为回文串,那么该字符必须是原字符串中的某一字符

故枚举原字符串中的所有字符,并枚举插入的位置,得到一个新串,判断是否是回文串即可.

import java.security.PublicKey;
import java.util.Scanner;
public class problem1
{
    public static void main(String[] args)
    {
        String s;
        Scanner scanner=new Scanner(System.in);
        s=scanner.nextLine();
        boolean flag=false;
        for(int i=0;i<s.length();i++)
        {
            if(flag)
                break;
            String str;
            for(int j=0;j<=s.length();j++)
            {
                str=s.substring(0,j)+s.charAt(i)+s.substring(j,s.length());
                if(judget(str)&&!flag)
                {
                    System.out.println(str);
                    flag=true;
                }
            }
        }
        if(!flag)
            System.out.println("NA");
    }
    public static boolean judget(String s)
    {
        for(int i=0;i<s.length();i++)
            if(s.charAt(i)!=s.charAt(s.length()-1-i))
                return false;
        return true;
    }
}