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;
int gcd(int a, int b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long digitcount(long long n) {
  long long c = 0;
  while (n > 0) {
    c++;
    n /= 10;
  }
  return c;
}
long long getndigit(long long n, long long a) {
  long long v = 0;
  while (n--) {
    v = v * 10 + a;
  }
  return v;
}
long long fx(long long x) {
  if (x < 10) {
    return x;
  }
  long long ans = 9;
  long long z = x, i = 0;
  while (x > 99) {
    ans += pow(10, i) * 9;
    i++;
    x /= 10;
  }
  ans += ((x / 10) - 1) * pow(10, i) + (z % 10 >= (x / 10));
  x = z;
  x /= 10;
  i = 0;
  while (x > 9) {
    ans += pow(10, i) * (x % 10);
    i++;
    x /= 10;
  }
  return ans;
}
int forw[500004][26], back[26][26];
string s[500004];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  memset(forw, 0, sizeof(forw));
  memset(back, 0, sizeof(back));
  int n, f, b, l;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    f = s[i].front() - 'a';
    b = s[i].back() - 'a';
    l = s[i].length();
    forw[i][f] = l;
    for (int j = 0; j < 26; j++) {
      if (back[f][j]) {
        forw[i][j] = max(forw[i][j], back[f][j] + l);
      }
      back[b][j] = max(back[b][j], forw[i][j]);
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    b = s[i].back() - 'a';
    ans = max(ans, forw[i][b]);
  }
  cout << ans;
  return 0;
}
