Problem: Given a string s, determine if it contains any palindrome of length exactly 100 as a subsequence. If it has any, print any one of them. If it doesn't have any, print a palindrome that is a subsequence of s and is as long as possible.

Input

The only line of the input contains one string s of length n (1 ≤ n ≤ 5·104) containing only lowercase English letters.

Output

If s contains a palindrome of length exactly 100 as a subsequence, print any palindrome of length 100 which is a subsequence of s. If s doesn't contain any palindromes of length exactly 100, print a palindrome that is a subsequence of s and is as long as possible.

If there exists multiple answers, you are allowed to print any of them.

Examples

Input

bbbabcbbb


Output

bbbcbbb


Input

rquwmzexectvnbanemsmdufrg


Output

rumenanemur

Note

A subsequence of a string is a string that can be derived from it by deleting some characters without changing the order of the remaining characters. A palindrome is a string that reads the same forward or backward.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char a[50011];
char answer[51];
char answer2[51];
int dp[2610][2610];
int dp2[2610][2610];
int main() {
  int len;
  scanf("%s", a + 1);
  len = strlen(a + 1);
  if (len >= 2600) len = 2600;
  int i, j, k, s, l, p;
  memset(dp, 0, sizeof(dp));
  memset(dp2, 0, sizeof(dp2));
  for (j = 0; j <= len; j++) {
    for (i = 1; j + i <= len; i++) {
      k = i + j;
      if (a[k] == a[i]) {
        if (i != k)
          dp[i][k] = dp[i + 1][k - 1] + 2;
        else {
          dp[i][k] = 1;
        }
        dp2[i][k] = 2;
      } else {
        if (dp[i + 1][k] > dp[i][k - 1]) {
          dp[i][k] = dp[i + 1][k];
          dp2[i][k] = 1;
        } else {
          dp[i][k] = dp[i][k - 1];
          dp2[i][k] = -1;
        }
      }
      if (dp[i][k] >= 100) break;
    }
    if (dp[i][k] >= 100) break;
  }
  j = 0;
  p = dp[i][k];
  while (i <= k) {
    if (dp2[i][k] == 2) {
      answer[j++] = a[k];
      i++;
      k--;
    }
    if (dp2[i][k] == 1) i++;
    if (dp2[i][k] == -1) k--;
  }
  l = 0;
  if (i - k == 2)
    s = j - 2;
  else
    s = j - 1;
  for (; s >= 0; s--) answer2[l++] = answer[s];
  answer2[l] = 0;
  if (p > 100)
    answer[j - 1] = 0;
  else
    answer[j] = 0;
  printf("%s", answer);
  printf("%s\n", answer2);
  return 0;
}