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 ofs. 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.

解题说明：此题是一道回文判断题，先按逆序保存字符串，然后再判断添加什么字母构成回文。

#include<cstdio>#include <cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>using namespace std;int main(){    char s[11],r[12];int l,i,j;    scanf("%s",s);    l=strlen(s);    for(i=0;i<=l;i++)    {        if(i<(l/2+1))        {r[i]=s[l-1-i];}else {r[i]=s[l-i];}        for(j=0;j<i;j++)        {r[j]=s[j];}for(j=i+1;j<=l;j++)        {r[j]=s[j-1];}   r[l+1]='\0';        for(j=0;j<(l/2+1);j++)        {            if(r[j]!=r[l-j]){break;}        }        if(j==(l/2+1)){break;}    }    if(i==l+1){printf("NA\n");}else {printf("%s\n",r);}return 0;}