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

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
int n;
string s;
int dp[28][28];
int ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j;
  int first, last;
  int sz;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> s;
    first = s[0] - 'a' + 1;
    last = s[s.size() - 1] - 'a' + 1;
    sz = s.size();
    if (dp[last][first] != 0) ans = max(ans, dp[last][first] + sz);
    if (first == last) ans = max(ans, sz);
    for (j = 1; j <= 26; j++)
      if (dp[j][first] != 0) dp[j][last] = max(dp[j][last], dp[j][first] + sz);
    dp[first][last] = max(dp[first][last], sz);
  }
  cout << ans << endl;
  return 0;
}
