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;
const int INF = (int)1e10;
string s;
int dp[5005][5005];
string res;
void trace(int x, int y) {
  if (x > y) return;
  if (x == y) {
    res += s[x];
    return;
  }
  if (s[x] == s[y]) {
    res += s[x];
    trace(x + 1, y - 1);
    res += s[y];
    return;
  }
  if (dp[x][y - 1] > dp[x + 1][y]) {
    trace(x, y - 1);
    return;
  } else {
    trace(x + 1, y);
    return;
  }
}
int rec(int x, int y) {
  if (x > y) return 0;
  if (dp[x][y]) return dp[x][y];
  if (x == y) return dp[x][y] = 1;
  if (s[x] == s[y]) return dp[x][y] = rec(x + 1, y - 1) + 2;
  return dp[x][y] = max(rec(x + 1, y), rec(x, y - 1));
}
void solve() {
  cin >> s;
  int n = s.size();
  if (n >= 2600) {
    map<char, int> m;
    for (int i = 0; i < s.size(); i++) {
      m[s[i]]++;
      if (m[s[i]] == 100) {
        for (int j = 1; j <= 100; j++) {
          cout << s[i];
        }
        return;
      }
    }
  }
  rec(0, n - 1);
  trace(0, n - 1);
  int t = res.size();
  if (t <= 100) {
    cout << res;
  } else {
    for (int i = 0; i < 50; i++) {
      cout << res[i];
    }
    for (int i = t - 50; i < res.size(); i++) {
      cout << res[i];
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
