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.
#include <bits/stdc++.h>
using namespace std;
const int N = 2610;
int n;
int fr[30];
int lcs[N][N];
string Ans, s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  for (int i = 0; i < s.size(); i++) fr[s[i] - 'a']++;
  for (int i = 0; i < 26; i++) {
    if (fr[i] > 99) {
      for (int j = 1; j <= 100; j++) cout << (char)(i + 'a');
      return 0;
    }
  }
  int n = s.size();
  string t = s;
  reverse(t.begin(), t.end());
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (!i || !j)
        lcs[i][j] = 0;
      else if (s[i - 1] == t[j - 1])
        lcs[i][j] = 1 + lcs[i - 1][j - 1];
      else
        lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]);
    }
  }
  int I = n, J = n, len = lcs[n][n];
  while (I && J) {
    if (s[I - 1] == t[J - 1]) {
      Ans += s[I - 1];
      I--;
      J--;
      len--;
    } else if (lcs[I - 1][J] == len)
      I--;
    else
      J--;
  }
  if (Ans.size() > 100) {
    if (Ans.size() % 2) Ans.erase(Ans.size() / 2, 1);
    int rem = Ans.size() - 100;
    rem /= 2;
    for (int i = rem; i < Ans.size() - rem; i++) cout << Ans[i];
  } else
    cout << Ans << "\n";
}
