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;
const int N = 2605;
int n, dp[N][N];
string s, ans;
void rec(int l, int r) {
  if (l > r) {
    return;
  }
  if (l == r) {
    ans += s[l];
  } else if (s[l] == s[r]) {
    ans += s[l];
    rec(l + 1, r - 1);
    ans += s[r];
  } else {
    if (dp[l + 1][r] >= dp[l][r - 1]) {
      rec(l + 1, r);
    } else {
      rec(l, r - 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = (int)s.size();
  if (n > 2600) {
    map<char, int> q;
    for (auto it : s) {
      q[it]++;
      if (q[it] == 100) {
        return cout << string(100, it), 0;
      }
    }
  } else {
    s = '#' + s;
    for (int sz = 1; sz <= n; sz++) {
      for (int l = 1; l + sz - 1 <= n; l++) {
        int r = l + sz - 1;
        if (l == r) {
          dp[l][r] = 1;
        } else 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]);
        }
      }
    }
    rec(1, n);
    if ((int)ans.size() <= 100) {
      cout << ans;
    } else {
      for (int i = 0; i < 50; i++) {
        cout << ans[i];
      }
      for (int i = (int)ans.size() - 50; i < (int)ans.size(); i++) {
        cout << ans[i];
      }
    }
  }
}