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;
int dp[3000][3000];
char s[60000];
int c[128];
int main(void) {
  scanf("%s", s);
  for (int i = 0; s[i]; i++) c[(int)s[i]]++;
  int n = strlen(s);
  for (int i = 'a'; i <= 'z'; i++)
    if (c[i] >= 100) {
      for (int j = 0; j < 100; j++) printf("%c", i);
      puts("");
      return 0;
    }
  for (int i = 0; i < n; i++) dp[1][i] = 1;
  for (int i = 0; i < n; i++)
    if (s[i] == s[i + 1])
      dp[2][i] = 2;
    else
      dp[2][i] = 1;
  for (int len = 3; len <= n; len++) {
    for (int i = 0; i + len <= n; i++)
      if (s[i] == s[i + len - 1])
        dp[len][i] = dp[len - 2][i + 1] + 2;
      else
        dp[len][i] = max(dp[len - 1][i], dp[len - 1][i + 1]);
  }
  int best = 0, bx = 0, by = 0;
  for (int len = 1; len <= n; len++)
    for (int i = 0; i < n; i++)
      if (best < dp[len][i]) {
        best = dp[len][i];
        bx = len;
        by = i;
      }
  char ans[3000] = {}, bns[3000] = {};
  int la = 0, lb = 0;
  while (bx > 0) {
    if (bx == 1) {
      ans[la++] = s[by];
      break;
    }
    if (s[by] == s[by + bx - 1]) {
      ans[la++] = s[by];
      bns[lb++] = s[by + bx - 1];
      bx -= 2;
      ++by;
    } else {
      if (dp[bx][by] == dp[bx - 1][by])
        --bx;
      else if (dp[bx][by] == dp[bx - 1][by + 1])
        --bx, ++by;
    }
  }
  if (la + lb >= 100) la = lb = 50;
  reverse(bns, bns + lb);
  ans[la] = 0;
  bns[lb] = 0;
  printf("%s%s\n", ans, bns);
  return 0;
}