cf 505A

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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
using namespace std;
string s;
bool check(string x)
{
      int len=x.size();
      for(int i=0;i<x.size();i++)
            if(x[i]!=x[len-1-i])
                  return false;
      return true;
}
int main()
{
      cin>>s;
      for(int i=0;i<=s.size();i++)
      {
            for(char k='a';k<='z';k++)
            {
                  string m=s.substr(0,i)+k+s.substr(i,s.size()-i);
                  if(check(m))
                  {
                        cout<<m<<endl;
                        return 0;
                  }
            }
      }
      cout<<"NA"<<endl;
      return 0;
}

  

posted @ 2015-01-21 20:02  心田定则  阅读(114)  评论(0编辑  收藏  举报