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 = 3000;
const int MOD = 1000000007;
string a;
int dp[N][N];
int par[N][N];
int chnum[26];
string get(int x, int y) {
  string ans;
  if (y - x == 1) return ans = a[x];
  if (par[x][y] == 0) {
    ans = a[x];
    ans += get(x + 1, y - 1);
    ans += a[y - 1];
    return ans;
  } else if (par[x][y] == 1)
    return get(x, y - 1);
  else if (par[x][y] == 2)
    return get(x + 1, y);
  return "";
}
int main() {
  cin >> a;
  memset(chnum, 0, sizeof(chnum));
  int n = (int)a.size();
  if (n >= 2600) {
    for (int i = 0; i < n; i++) chnum[a[i] - 'a']++;
    for (int i = 0; i < 26; i++) {
      if (chnum[i] >= 100) {
        for (int j = 0; j < 100; j++) cout << (char)(i + 'a');
        cout << endl;
        return 0;
      }
    }
  }
  for (int i = n; i >= 0; i--) {
    for (int j = 0; j <= n; j++) {
      if (j - i < 2) {
        if (j - i == 1) dp[i][j] = 1;
        par[i][j] = -1;
        continue;
      }
      if (a[i] == a[j - 1]) {
        dp[i][j] = dp[i + 1][j - 1] + 2;
        par[i][j] = 0;
      } else {
        dp[i][j] = max(dp[i][j - 1], dp[i + 1][j]);
        if (dp[i][j] == dp[i][j - 1])
          par[i][j] = 1;
        else
          par[i][j] = 2;
      }
    }
  }
  string s = get(0, n);
  if (s.size() <= 100)
    cout << s << endl;
  else {
    cout << s.substr(0, 50);
    cout << s.substr((int)s.size() - 50) << endl;
  }
  return 0;
}