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;
map<char, int> cnt;
string s, res;
int dp[3000][3000];
void get(int l, int r) {
  if (l > r || res.size() == 50) return;
  if (s[l] == s[r]) {
    res += s[l];
    get(l + 1, r - 1);
  } else if (dp[l + 1][r] > dp[l][r - 1])
    get(l + 1, r);
  else
    get(l, r - 1);
}
int main() {
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) cnt[s[i]]++;
  for (auto &it : cnt)
    if (it.second >= 100) {
      for (int i = 0; i < 100; i++) cout << it.first;
      return 0;
    }
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int d = 1; d < n; d++) {
    for (int j = 0; j + d < n; j++) {
      int l = j, r = j + d;
      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]);
    }
  }
  get(0, n - 1);
  cout << res;
  reverse(res.begin(), res.end());
  if (dp[0][n - 1] % 2 && dp[0][n - 1] < 100) res.erase(0, 1);
  cout << res;
}