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;
int n;
string ans;
int dp[2600][2600];
int cnt[27];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = (int)s.size();
  for (int i = 0; i < n; i++) {
    cnt[s[i] - 'a']++;
    if (cnt[s[i] - 'a'] >= 100) {
      for (int j = 1; j <= 100; j++) {
        cout << s[i];
      }
      return 0;
    }
  }
  for (int i = 0; i <= n; i++) dp[i][i] = 1;
  for (int len = 2; len <= n; len++) {
    for (int l = 0; l + len - 1 < n; l++) {
      int r = l + len - 1;
      if (s[l] == s[r]) {
        dp[l][r] = 2 + dp[l + 1][r - 1];
      } else {
        if (dp[l][r - 1] > dp[l + 1][r])
          dp[l][r] = dp[l][r - 1];
        else
          dp[l][r] = dp[l + 1][r];
      }
    }
  }
  int l = 0, r = n - 1;
  while (1) {
    if (l > r || ans.size() == 50) break;
    if (s[l] == s[r]) {
      ans += s[l];
      l++;
      r--;
    } else if (dp[l + 1][r] > dp[l][r - 1])
      l++;
    else
      r--;
  }
  cout << ans;
  reverse(ans.begin(), ans.end());
  if (dp[0][n - 1] % 2 && dp[0][n - 1] < 100) ans.erase(0, 1);
  cout << ans;
  return 0;
}
