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;
int kol[100];
pair<int, int> xx, yy;
int ans;
int dp[2606][2606];
pair<int, int> pred[2606][2606];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  if (s.size() >= 2600) {
    for (int i = 0; i < s.size(); ++i) ++kol[s[i] - 'a'];
    for (int i = 0; i < 26; ++i)
      if (kol[i] >= 100) {
        for (int j = 0; j < 100; ++j) cout << char(i + 'a');
        return 0;
      }
  }
  int n = s.size();
  xx = yy = make_pair(-1, -1);
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j <= n - i + 1; ++j) {
      pred[j][j + i - 1] = make_pair(j, j + i - 2);
      dp[j][j + i - 1] = dp[j][j + i - 2];
      if (dp[j + 1][j + i - 1] > dp[j][j + i - 1]) {
        dp[j][j + i - 1] = dp[j + 1][j + i - 1];
        pred[j][j + i - 1] = make_pair(j + 1, j + i - 1);
      }
      if (s[j - 1] == s[j + i - 2] &&
          dp[j + 1][j + i - 2] + 1 > dp[j][j + i - 1]) {
        dp[j][j + i - 1] = dp[j + 1][j + i - 2] + 1;
        pred[j][j + i - 1] = make_pair(j + 1, j + i - 2);
      }
      if (dp[j][j + i - 1] == 1 && s[j - 1] == s[j + i - 2])
        pred[j][j + i - 1] = make_pair(-1, -1);
      if (dp[j][j + i - 1] > ans) {
        ans = dp[j][j + i - 1];
        xx = make_pair(j, j + i - 1);
      }
      if (dp[j][j + i - 1] == 50) {
        yy = make_pair(j, j + i - 1);
      }
    }
  }
  if (xx == make_pair(-1, -1)) {
    cout << s[0];
    return 0;
  }
  if (yy != make_pair(-1, -1)) {
    xx = yy;
    string res = "";
    while (xx != make_pair(-1, -1)) {
      if (xx.first == xx.second) break;
      if (pred[xx.first][xx.second] == make_pair(-1, -1) ||
          dp[xx.first][xx.second] == dp[pred[xx.first][xx.second].first]
                                       [pred[xx.first][xx.second].second] +
                                         1)
        res += s[xx.first - 1];
      xx = pred[xx.first][xx.second];
    }
    string res2 = res;
    reverse(res2.begin(), res2.end());
    if (xx != make_pair(-1, -1) && xx.first == xx.second)
      res += s[xx.first - 1];
    res += res2;
    cout << res;
    return 0;
  }
  xx = yy = make_pair(-1, -1);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n - i + 1; ++j) {
      pred[j][j + i - 1] = make_pair(j, j + i - 2);
      dp[j][j + i - 1] = dp[j][j + i - 2];
      if (dp[j + 1][j + i - 1] > dp[j][j + i - 1]) {
        dp[j][j + i - 1] = dp[j + 1][j + i - 1];
        pred[j][j + i - 1] = make_pair(j + 1, j + i - 1);
      }
      if (s[j - 1] == s[j + i - 2] &&
          dp[j + 1][j + i - 2] + (i == 1 ? 1 : 2) > dp[j][j + i - 1]) {
        dp[j][j + i - 1] = dp[j + 1][j + i - 2] + (i == 1 ? 1 : 2);
        pred[j][j + i - 1] = make_pair(j + 1, j + i - 2);
      }
      if (dp[j][j + i - 1] == 1 && s[j - 1] == s[j + i - 2])
        pred[j][j + i - 1] = make_pair(-1, -1);
      if (dp[j][j + i - 1] > ans) {
        ans = dp[j][j + i - 1];
        xx = make_pair(j, j + i - 1);
      }
    }
  }
  string res = "";
  bool t = false;
  if (dp[xx.first][xx.second] & 1) t = true;
  while (xx != make_pair(-1, -1)) {
    if (xx.first == xx.second) break;
    if (pred[xx.first][xx.second] == make_pair(-1, -1) ||
        dp[xx.first][xx.second] == dp[pred[xx.first][xx.second].first]
                                     [pred[xx.first][xx.second].second] +
                                       2)
      res += s[xx.first - 1];
    xx = pred[xx.first][xx.second];
  }
  string res2 = res;
  reverse(res2.begin(), res2.end());
  if (t) res += s[xx.first - 1];
  res += res2;
  cout << res;
}