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;
int N, M, T;
int dp[127];
int tb[2600][2600];
int main() {
  int i, j, k;
  string s1, s2;
  cin >> s1;
  if (s1.size() >= 2575) {
    for (i = 0; i < s1.size(); i++) dp[s1[i]]++;
    for (i = 'a'; i <= 'z'; i++)
      if (dp[i] >= 100) break;
    for (j = 0; j < 100; j++) printf("%c", i);
    return 0;
  }
  s2 = s1;
  reverse((s2).begin(), (s2).end());
  N = s1.size();
  for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
      tb[i + 1][j + 1] = max(tb[i][j + 1], tb[i + 1][j]);
      if (s1[i] == s2[j]) {
        tb[i + 1][j + 1] = max(tb[i + 1][j + 1], tb[i][j] + 1);
      }
    }
  }
  i = N, j = N;
  string ans = "";
  while (i > 0 && j > 0) {
    if (s1[i - 1] == s2[j - 1] && tb[i - 1][j - 1] + 1 == tb[i][j]) {
      ans += s1[i - 1];
      i--;
      j--;
    } else if (tb[i - 1][j] == tb[i][j])
      i--;
    else
      j--;
  }
  if (ans.size() > 100) {
    string ss = ans;
    ans = "";
    for (i = 0; i < 50; i++) ans += ss[i];
    ss = ans;
    reverse((ss).begin(), (ss).end());
    ans += ss;
  }
  cout << ans;
}