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

import sys
# for fast input
#setting up
n= int(sys.stdin.readline())
s = []
# list of string
for i in range (n):
     temp = sys.stdin.readline()
     s.append(temp)
f= [[-1 for i in range (123)] for x in range (123)]
# interger memoization array f[i][j] is the longest sequence name that pass all the constrain and end with  chr (i) and start with chr (j)
# i and j are ord for the last and first letter

#dp starts here  --------------------------------------------------------------------------------------------------------

for i in s :
    # first we look through all the name in s , i is now a string in s
    # first , we develop the memoization array f with i
    for j in range (97,123):
        if f[ord (i[0])][j] != -1:
        #first I want to assume a seq start with chr j and end with i's first is available
             f[ord (i[len(i)-2])][j] = max(f[ord (i[0])][j] + len(i)-1,f[ord (i[len(i)-2])][j])
             # we gonna optimize following this formula:
             # length of a seq start with chr (j) and end i's last letter
             # = len of seq start with chr j and end with i's first + len of i (as we add the string we're considering to the seq)
    # after that  we assign the string i to f[i][j]
    f[ord(i[len(i)-2])][ord(i[0])]  = max (f[ord(i[len(i)-2])][ord(i[0])], len (i)-1)
    # noitice that we avoid assigning first as it may cause a Repetition
#dp ends here-----------------------------------------------------------------------------------------------------------

# now we  choose the result, the result seq must start and end with the same character
res = 0
for i in range (97,123):
    res = max (res,f[i][i])
print (res)

