This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.

You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.

We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤  p1 < p2 < ... < pk ≤  |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".

String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".

String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".

Input

The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.

Output

Print the lexicographically largest palindromic subsequence of string s.

Examples

Input

radar


Output

rr


Input

bowwowwow


Output

wwwww


Input

codeforces


Output

s


Input

mississipp


Output

ssss

Note

Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
import bisect
import decimal
from decimal import Decimal
import os
from collections import Counter
import bisect
from collections import defaultdict
import math
import random
import heapq 
from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations
from io import BytesIO, IOBase
from itertools import accumulate

# sys.setrecursionlimit(200000)
# mod = 10**9+7
# mod = 998244353
decimal.getcontext().prec = 46

def primeFactors(n): 
    prime = set()
    while n % 2 == 0: 
        prime.add(2)
        n = n//2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            prime.add(i)
            n = n//i 
    if n > 2: 
        prime.add(n)
    return list(prime)
    
def getFactors(n) : 
    factors = []
    i = 1
    while i <= math.sqrt(n): 
        if (n % i == 0) : 
            if (n // i == i) : 
                factors.append(i)
            else : 
                factors.append(i)
                factors.append(n//i)
        i = i + 1
    return factors
    
def SieveOfEratosthenes(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    num = []
    for p in range(2, n+1): 
        if prime[p]: 
            num.append(p)
    return num

def lcm(a,b):
    return (a*b)//math.gcd(a,b)

def sort_dict(key_value):
    return sorted(key_value.items(), key = lambda kv:(kv[1], kv[0]))

def list_input():
    return list(map(int,input().split()))
    
def num_input():
    return map(int,input().split())
    
def string_list():
    return list(input())
    
def decimalToBinary(n):  
    return bin(n).replace("0b", "")  
    
def binaryToDecimal(n): 
    return int(n,2) 
    
def solve():
    s = input()
    occur = {}
    for i in s:
        try:
            occur[i] += 1
        except KeyError:
            occur[i] = 1
    s = list(occur.keys())
    maxx = max(s)
    print(maxx*occur[maxx])
    
#t = int(input())
t = 1
for _ in range(t):
    solve()


    
    
    
    
