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;
int p[2582][2582];
string a;
int tam(int i, int j) {
  if (p[i][j] != -1) return p[i][j];
  if (i == j) return p[i][j] = 1;
  if (i + 1 == j and a[i] == a[j]) return p[i][j] = 2;
  if (a[i] == a[j]) return p[i][j] = 2 + tam(i + 1, j - 1);
  return p[i][j] = max(tam(i + 1, j), tam(i, j - 1));
}
string rec(int i, int j) {
  if (i == j) return string() + a[i];
  if (i + 1 == j) {
    if (a[i] == a[j]) return string() + a[i] + a[j];
    return string() + a[i];
  }
  string r;
  if (a[i] == a[j])
    r = string() + a[i] + rec(i + 1, j - 1) + a[j];
  else if (p[i + 1][j] > p[i][j - 1])
    r = rec(i + 1, j);
  else
    r = rec(i, j - 1);
  return r;
}
int main() {
  cin >> a;
  if (a.size() > 2580) {
    a = a.substr(0, 2580);
  }
  memset(p, -1, sizeof(p));
  tam(0, a.size() - 1);
  a = rec(0, a.size() - 1);
  string u, w;
  if (a.size() > 100) {
    u = "";
    w = "";
    for (int i = 0; i < 50; i++) w += a[i];
    for (int i = 0; i < 50; i++) u += a[a.size() - i - 1];
    reverse(u.begin(), u.end());
    w = w + u;
  } else
    w = a;
  cout << w;
  return 0;
}