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 = 3010;
int dp[N][N], cnt[30];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) cnt[s[i] - 'a']++;
  for (int i = 0; i < 26; i++) {
    if (99 < cnt[i]) {
      for (int j = 0; j < 100; j++) cout << (char)('a' + i);
      return 0;
    }
  }
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int len = 1; len < n; len++) {
    for (int l = 0, r = len; r < n; l++, r++) {
      if (s[l] == s[r])
        dp[l][r] = dp[l + 1][r - 1] + 2;
      else
        dp[l][r] = max(dp[l + 1][r], dp[l][r - 1]);
    }
  }
  int l = 0, r = n - 1;
  string ans;
  while (l < r) {
    if (s[l] == s[r]) {
      ans += s[l];
      l++;
      r--;
    } else if (dp[l + 1][r] < dp[l][r - 1])
      r--;
    else
      l++;
  }
  for (int i = 0; i < min((int)(ans.size()), 50); i++) cout << ans[i];
  if (l == r && min((int)(ans.size()), 50) < 50) cout << s[l];
  for (int i = min((int)(ans.size()), 50) - 1; i > -1; i--) cout << ans[i];
  return 0;
}
