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 = 51000;
const int L = 26;
char s[N];
int cnt[L];
const int M = 2610;
char t[M];
short dp[M][M];
char pre[M][M];
char ans[M];
int k;
int main() {
  scanf("%s", s);
  int n = strlen(s);
  if (n > 2600) {
    for (int i = 0; i < n; i++) cnt[s[i] - 'a']++;
    for (int i = 0; i < L; i++)
      if (cnt[i] >= 100) {
        for (int j = 0; j < 100; j++) putchar('a' + i);
        puts("");
        return 0;
      }
    while (1)
      ;
  }
  for (int i = 0; i < n; i++) t[n - 1 - i] = s[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (s[i] == t[j]) {
        if (i && j)
          dp[i][j] = dp[i - 1][j - 1] + 1;
        else
          dp[i][j] = 1;
        pre[i][j] = 0;
      } else {
        dp[i][j] = 0;
        pre[i][j] = 1;
        if (i) {
          dp[i][j] = dp[i - 1][j];
          pre[i][j] = 1;
        }
        if (j && (!i || dp[i][j - 1] > dp[i - 1][j])) {
          dp[i][j] = dp[i][j - 1];
          pre[i][j] = 2;
        }
      }
    }
  }
  int x = n - 1, y = n - 1;
  k = 0;
  while (1) {
    if (pre[x][y] == 0) {
      ans[k++] = s[x];
      x--, y--;
    } else if (pre[x][y] == 1) {
      x--;
    } else {
      y--;
    }
    if (x < 0 || y < 0) break;
  }
  if (k < 100) {
    ans[k++] = 0;
    puts(ans);
  } else {
    for (int i = 0; i < 50; i++) putchar(ans[i]);
    for (int i = 50 - 1; i >= 0; i--) putchar(ans[i]);
    puts("");
  }
  return 0;
}