Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.

Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.

Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.

Output

In the first line, print the length of the longest palindrome string you made.

In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.

Examples

Input


3 3
tab
one
bat


Output


6
tabbat


Input


4 2
oo
ox
xo
xx


Output


6
oxxxxo


Input


3 5
hello
codef
orces


Output


0



Input


9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji


Output


20
ababwxyzijjizyxwbaba

Note

In the first example, "battab" is also a valid answer.

In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.

In the third example, the empty string is the only valid palindrome string.
#DaRk DeveLopeR
import sys


#taking input as string 
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
import math
from bisect import *
from heapq import *
from collections import defaultdict as dd
from collections import OrderedDict as odict
from collections import Counter as cc
from collections import deque
from itertools import groupby
sys.setrecursionlimit(20*20*20*20+10) #this is must for dfs



def solve():

	n,m=takeivr()


	dict_1=dd(lambda:0)

	for i in range(n):
		dict_1[takesr()]=1
	flag=False
	ans=""
	for i in list(dict_1.keys()):

		if ispalindrome(i):
			flag=i

		elif i[::-1] in dict_1 and dict_1[i[::-1]]:
			ans+=i
			dict_1[i]=0

	first=ans
	if flag!=False:
		first+=flag
	first+=ans[::-1]
	print(len(first))
	print(first)







def main():
    global tt
    if not ONLINE_JUDGE:
        sys.stdin = open("input.txt","r")
        sys.stdout = open("output.txt","w")
    t = 1
    # t = takein()
    #t = 1
    for tt in range(1,t + 1):
        solve()
    if not ONLINE_JUDGE:
        print("Time Elapsed :",time.time() - start_time,"seconds")
        sys.stdout.close()

#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def takein():
    return (int(sys.stdin.readline().rstrip("\r\n")))


# input the string

def takesr():
    return (sys.stdin.readline().rstrip("\r\n"))


# input int array
def takeiar():
    return (list(map(int, sys.stdin.readline().rstrip("\r\n").split())))


# input string array
def takesar():
    return (list(map(str, sys.stdin.readline().rstrip("\r\n").split())))


# innut values for the diffrent variables
def takeivr():
    return (map(int, sys.stdin.readline().rstrip("\r\n").split()))


def takesvr():
    return (map(str, sys.stdin.readline().rstrip("\r\n").split()))

 
 
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#



def ispalindrome(s):

	return s==s[::-1]
 
def invert(bit_s):


	  
	# convert binary string 
	# into integer
	temp = int(bit_s, 2)
	  
	# applying Ex-or operator
	# b/w 10 and 31
	inverse_s = temp ^ (2 ** (len(bit_s) + 1) - 1)
	  
	# convert the integer result 
	# into binary result and then 
	# slicing of the '0b1' 
	# binary indicator
	rslt = bin(inverse_s)[3 : ]

	return str(rslt) 	  

def counter(a):
    q = [0] * max(a)
    for i in range(len(a)):
        q[a[i] - 1] = q[a[i] - 1] + 1
    return(q)
 
def counter_elements(a):
    q = dict()
    for i in range(len(a)):
        if a[i] not in q:
            q[a[i]] = 0
        q[a[i]] = q[a[i]] + 1
    return(q)
 
def string_counter(a):
    q = [0] * 26
    for i in range(len(a)):
        q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
    return(q)
 
def factorial(n,m = 1000000007):
    q = 1
    for i in range(n):
        q = (q * (i + 1)) % m
    return(q)
 
def factors(n):
    q = []
    for i in range(1,int(n ** 0.5) + 1):
        if n % i == 0: q.append(i); q.append(n // i)
    return(list(sorted(list(set(q)))))
 
def prime_factors(n):
    q = []
    while n % 2 == 0: q.append(2); n = n // 2
    for i in range(3,int(n ** 0.5) + 1,2):
        while n % i == 0: q.append(i); n = n // i
    if n > 2: q.append(n)
    return(list(sorted(q)))
 
def transpose(a):
    n,m = len(a),len(a[0])
    b = [[0] * n for i in range(m)]
    for i in range(m): 
        for j in range(n): 
            b[i][j] = a[j][i]
    return(b)
 
def power_two(x):
    return (x and (not(x & (x - 1))))
 
def ceil(a, b):
    return -(-a // b)



def seive(n):
	a = [1]
	prime = [True for i in range(n+1)] 
	p = 2
	while (p * p <= n): 
	    if (prime[p] == True): 
	        for i in range(p ** 2,n + 1, p): 
	            prime[i] = False
	    p = p + 1
	for p in range(2,n + 1): 
	    if prime[p]: 
	        a.append(p)
	return(a)

 
def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1]+i)
    return pref_sum
 
def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far

def binary_search(li, val):
    # print(lb, ub, li)
    ans = -1
    lb = 0
    ub = len(li)-1
    while (lb <= ub):
        mid = (lb+ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid-1
        elif val > li[mid]:
            lb = mid+1
        else:
            ans = mid  # return index
            break
    return ans


 
def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li)-1
 
    while (start <= end):
        middle = (end+start) // 2
 
        if li[middle] <= num:
            answer = middle
            start = middle+1
 
        else:
            end = middle-1
    return answer  # max index where x is not greater than num


def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li)-1
 
    while (start <= end):
        middle = (end+start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle-1
        else:
            start = middle+1
    return answer  # min index where x is not less than num
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
    input = sys.stdin.readline
    
main()