/*

     题意：问最少插入多少个字符使得字符串变成回文串

     区间DP：dp[i][j]表示[l, r]的字符串要成为回文需要插入几个字符串，那么dp[l][r] = dp[l+1][r-1]; (str[l] == str[r])

             dp[l][r] = min (dp[l+1][r], dp[l][r-1]) + 1，然后按照状态转移递归输出路径

 */

 /************************************************

 * Author        :Running_Time

 * Created Time  :2015-8-17 14:42:57

 * File Name     :UVA_10453.cpp

  ************************************************/

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <sstream>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <vector>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <list>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cstdlib>

 #include <ctime>

 using namespace std;

 #define lson l, mid, rt << 1

 #define rson mid + 1, r, rt << 1 | 1

 typedef long long ll;

 const int MAXN = 1e3 + ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 char str[MAXN];

 int dp[MAXN][MAXN];

 void print(int l, int r)    {

     if (l > r)  return ;

     if (l == r) printf ("%c", str[l]);

     else if (str[l] == str[r])   {

         printf ("%c", str[l]);

         print (l + , r - );

         printf ("%c", str[l]);

     }

     else if (dp[l][r] == dp[l+][r] + )    {

         printf ("%c", str[l]);

         print (l + , r);

         printf ("%c", str[l]);

     }

     else    {

         printf ("%c", str[r]);

         print (l, r - );

         printf ("%c", str[r]);

     }

 }

 void work(void) {

     memset (dp, , sizeof (dp));

     int len = strlen (str);

     for (int i=; i<=len; ++i)  {

         for (int j=; j+i-<len; ++j)   {

             int k = j + i - ;

             int &res = dp[j][k] = INF;

             if (str[j] == str[k])   res = dp[j+][k-];

             res = min (res, min (dp[j+][k], dp[j][k-]) + );

         }

     }

     printf ("%d ", dp[][len-]);

     print (, len - ); puts ("");

 }

 int main(void)    {     //UVA 10453 Make Palindrome

     while (scanf ("%s", str) == )  {

         work ();

     }

     return ;

 }