10
abdcced
bdacdac
ecb
dc
eaeeebdd
adcdbadcac
eb
aadecccde
edbaeacad
accd
 100
d
a
e
eb
c
d
d
e
c
e
ac
d
c
aa
a
e
da
e
e
c
e
a
c
c
bc
bd
d
da
cc
be
b
a
ee
d
be
be
e
b
a
ed
cb
ab
a
ea
ac
aa
ed
d
b
eb
e
ed
c
ce
c
d
c
d
d
ee
e
b
ea
ea
da
ab
ba
ed
da
ec
ea
bd
ee
dc
d
bd
d
c
cc
ab
bd
a
ad
e
d
e
bc
bb
c
bc
a
a
e
ca
ac
c
ad
c
be
d
 3
ab
acb
ba
 10
a
b
a
c
b
a
b
a
a
a
 5
cab
bbc
caa
ccc
aca
 3
ac
abc
ca
 5
ab
bc
cd
de
ffffffffff
 16
ddgcb
fgbhgbdg
gha
dh
ddfb
fggbc
edb
hd
d
hfgcabf
hdaef
fhbcdgba
chee
cafeggc
f
ecb
 50
agecd
cghafi
fiide
iecc
jbdcfjhgd
diee
hfeg
ehc
gfijgjh
gacaifebg
dicbbddc
hjgciaei
jjcdh
g
gc
cf
hfdjhd
c
icidbec
ji
eeh
cgeejggc
acfd
jjg
ajefdj
eghhebiic
bih
gbb
jjaa
idc
gafi
ecg
dbigbjiehj
cedif
ahidfaaajg
hfhhiccbgb
dgegjgieif
hgjebhfdc
j
ahehd
ahdah
bijbhh
a
dabiggfhac
dedigbjbf
ihbdj
iihagchd
gfebhfdi
jbjdcggdh
f

0
 35
 5
 6
 9
 5
 10
 7
 37

//package Test_A;

import java.util.Arrays;
import java.util.Scanner;

//这道题目说的就是要在已知的两个条件的约束下达到满足这个的最大的长度
//一定要注意是在满足两个条件的情况下的成立 
//这道题目就是简单的DP  O(∩_∩)O哈哈~   明白了
//就是枚举以每个为开始的最大的长度   最后比较出最长的就好了
public class Main {
    static int n ;
    static String a[] = new String [500010];
    static int dp[][] = new int [30][30];
    static char str[] = new char[30];
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		n = scan.nextInt();
		int length = 0;
		int l = 0;
		int r = 0;
		for(int i=0;i<30;i++)
			for(int j=0;j<30;j++)
				dp[i][j] = 0;
		   for(int k=1;k<=n;k++)
		    {  
		       String string  = scan.next();  
		       str = string.toCharArray();
		       length =string.length();  
		        l = str[0] - 'a';  
		        r = str[length-1] - 'a';  
		        for(int i=0;i<26;i++)  
		            if(dp[i][l]!=0)  
		                dp[i][r] = Math.max(dp[i][r],dp[i][l]+length);  
		        dp[l][r] = Math.max(dp[l][r],length);  
		    }  
		    int ans = 0;  
		    for(int i=0;i<26;i++)  
		        ans = Math.max(ans,dp[i][i]);  
		   System.out.println(ans);
		}  
	}


 				  	 				  		 	   				