Mr. Kitayuta's Gift

B - Mr. Kitayuta's Gift

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

SubmitStatusPractice CodeForces 505A uDebug

Description

Input

Output

Sample Input

Sample Output

Hint

Description

Mr. Kitayuta has kindly given you a string s consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter intos 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 ins 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 Input

Input

revive

Output

reviver

Input

ee

Output

eye

Input

kitayuta

Output

NA

Sample Output

Hint

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

思路：暴力解决

26个字母，11个空，挨个插进去

#include<cstdio>#include<cstring>using namespace std;const int N = 20;char s[N], p[N];int l;bool huiwen(){    for(int i = 0; i < (l + 1) / 2; i++)        if(p[i] != p[l - i]) return false;    return true;}int main(){    int i, j, k;    while(scanf("%s", s)!=-1)    {    l = strlen(s);    for(char c = 'a'; c <= 'z'; ++c)    {        for(k = 0; k <= l; ++k)        {            i = j = -1;            while(i < k - 1) p[++j] = s[++i];            p[++j] = c;            while(i < l - 1) p[++j] = s[++i];            if(huiwen())            {            printf("%s\n", p);             return 0;            }        }    }    printf("NA\n");    }    return 0;}