human
Alice bought a Congo Prime Video subscription and was watching a documentary on the archaeological findings from Factor's Island on Loch Katrine in Scotland. The archaeologists found a book whose age and origin are unknown. Perhaps Alice can make some sense of it?

The book contains a single string of characters "a", "b" and "c". It has been pointed out that no two consecutive characters are the same. It has also been conjectured that the string contains an unusually long subsequence that reads the same from both sides. 

Help Alice verify this by finding such subsequence that contains at least half of the characters of the original string, rounded down. Note that you don't have to maximise the length of it.

A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.

Input

The input consists of a single string s (2 ≤ |s| ≤ 10^6). The string s consists only of characters "a", "b", "c". It is guaranteed that no two consecutive characters are equal.

Output

Output a palindrome t that is a subsequence of s and |t| ≥ ⌊ (|s|)/(2) ⌋.

If there are multiple solutions, you may print any of them. You don't have to maximise the length of t.

If there are no solutions, output a string "IMPOSSIBLE" (quotes for clarity).

Examples

Input

cacbac

Output

aba

Input

abc

Output

a

Input

cbacacacbcbababacbcb

Output

cbaaacbcaaabc

Note

In the first example, other valid answers include "cacac", "caac", "aca" and "ccc". 
Using python can you solve the prior task?
assistant
#Here is Python code to print any of them.

from __future__ import division, print_function
def main():
    s=input()
    i=0
    j=len(s)-1
    ans=[]
    while i+1<j-1:
        if s[i]==s[j]:
            ans.append(s[i])
        elif s[i]==s[j-1]:
            ans.append(s[i])
        elif s[i+1]==s[j]:
            ans.append(s[i+1])
        elif s[i+1]==s[j-1]:
            ans.append(s[i+1])
        i+=2
        j-=2
    if len(s)%4!=0:
        ans.append(s[i])
        temp=[]
        for c in ans:
            temp.append(c)
        temp.pop()
        temp=temp[::-1]
    else:
        temp=[]
        for c in ans:
            temp.append(c)
        temp=temp[::-1]
    ans.extend(temp)
    print(''.join(ans))
        
            
        
        
                
######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO,...