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;
char str[51234];
int L[51234][123];
int P[51234][27];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> str;
  n = strlen(str);
  for (int i = 0; i < 26; ++i) {
    int w = n;
    P[n][i] = 0x3f3f3f3f;
    for (int j = n - 1; j >= 0; --j) {
      if (str[j] - 'a' == i) w = j;
      P[j][i] = w;
    }
  }
  for (int i = 0; i <= n; ++i) {
    L[i][0] = i - 1;
    L[i][1] = i;
  }
  memset(L, 0x3f, sizeof(L));
  for (int i = 0; i < n; ++i) {
    L[i][0] = i - 1;
    L[i][1] = i;
  }
  int mx = 1;
  for (int k = 2; k < 102; ++k) {
    for (int i = n - 1; i >= 0; --i) {
      if (i < n) {
        int t = L[i + 1][k - 2] + 1;
        if (t < n) {
          L[i][k] = min(L[i + 1][k], P[t][str[i] - 'a']);
        } else {
          L[i][k] = L[i + 1][k];
        }
      }
    }
    if (L[0][k] < n) mx = k;
  }
  int l, r;
  bool flag = 0;
  for (int i = 0; i < n; ++i) {
    for (int k = 100; k >= 1; --k) {
      if (L[i][k] < n) {
        l = i, r = L[i][k];
        n = k;
        flag = 1;
      }
      if (flag) break;
    }
    if (flag) break;
  }
  string s1;
  for (int i = l; i <= r; ++i) {
    s1 += str[i];
  }
  string prefix, surfix;
  for (int i = 0; i < s1.length(); ++i) {
    if (L[i + 1 + l][n] < l + s1.length()) {
      s1 = s1.substr(0, L[i + 1 + l][n] - l + 1);
      continue;
    } else if (L[i + l][n] == l + s1.length() - 1) {
      prefix = prefix + s1[i];
      n--;
      if (i != s1.length() - 1) surfix = s1[s1.length() - 1] + surfix, n--;
      s1 = s1.substr(0, s1.length() - 1);
    } else {
      s1 = s1.substr(0, s1.length() - 1);
      i--;
    }
    if (n <= 0) break;
  }
  cout << prefix << surfix << endl;
  return 0;
}
