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;
string s;
const int maxn = 5100;
int dp[maxn][maxn];
int h[maxn];
string go(int l, int r, int sa) {
  if (sa == 100) return "";
  if (l > r) return "";
  if (l == r) {
    string tt = "";
    tt += s[l];
    return tt;
  }
  string t = "";
  if (dp[l][r] == dp[l + 1][r]) {
    t = go(l + 1, r, sa);
  } else if (dp[l][r] == dp[l][r - 1]) {
    t = go(l, r - 1, sa);
  } else {
    t += s[l];
    t += go(l + 1, r - 1, sa + 2);
    t += s[r];
  }
  return t;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    h[s[i]]++;
  }
  for (int i = 'a'; i <= 'z'; i++) {
    if (h[i] >= 100) {
      for (int j = 0; j < 100; j++) {
        cout << (char)(i);
      }
      return 0;
    }
  }
  for (int i = 0; i < s.size(); i++) {
    dp[i][i] = 1;
  }
  for (int len = 1; len < s.size(); len++) {
    int r = len;
    for (int l = 0; r < s.size(); l++, r++) {
      dp[l][r] = max(dp[l][r], dp[l + 1][r]);
      dp[l][r] = max(dp[l][r], dp[l][r - 1]);
      if (s[l] == s[r]) {
        dp[l][r] = max(dp[l][r], dp[l + 1][r - 1] + 2);
      }
    }
  }
  cout << go(0, s.size() - 1, 0) << endl;
  return 0;
}