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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int cnt[27];
int dp[3000][3000];
string s, ans;
int main() {
  cin >> s;
  int len = int(s.size());
  for (int i = 0; i < s.size(); i++) {
    cnt[s[i] - 'a']++;
  }
  for (int i = 0; i < 26; i++)
    if (cnt[i] > 100) {
      for (int j = 0; j < 100; j++) cout << char(i + 'a');
      cout << endl;
      return 0;
    }
  int i, j;
  for (i = 1; i <= s.size(); i++)
    for (j = 1; j <= s.size(); j++) {
      if (s[i - 1] == s[len - j])
        dp[i][j] = dp[i - 1][j - 1] + 1;
      else
        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
    }
  i--, j--;
  while (i > 0 && j > 0) {
    if (s[i - 1] == s[len - j]) {
      ans += s[i - 1];
      i--;
      j--;
    } else if (dp[i - 1][j] == dp[i][j])
      i--;
    else
      j--;
  }
  if (ans.size() > 100) {
    for (int i = 0; i < 50; i++) cout << ans[i];
    for (int i = 49; i >= 0; i--) cout << ans[i];
    cout << endl;
  } else
    cout << ans << endl;
  return 0;
}
