10
abdcced
bdacdac
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dc
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edbaeacad
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 100
d
a
e
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c
d
d
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c
e
ac
d
c
aa
a
e
da
e
e
c
e
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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
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c
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ab
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a
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d
e
bc
bb
c
bc
a
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ca
ac
c
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be
d
 3
ab
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 10
a
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a
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a
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 5
cab
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caa
ccc
aca
 3
ac
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ca
 5
ab
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cd
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ffffffffff
 16
ddgcb
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gha
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fggbc
edb
hd
d
hfgcabf
hdaef
fhbcdgba
chee
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f
ecb
 50
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gacaifebg
dicbbddc
hjgciaei
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f

0
 35
 5
 6
 9
 5
 10
 7
 37

#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using pi = pair<int, int>;
using ll = long long int;
template <typename T1, typename T2, typename T3>
T1 modpow(T1 _a, T2 p, T3 mod) {
  assert(p >= 0);
  ll ret = 1, a = _a;
  if (a < 0) {
    a %= mod;
    a += mod;
  }
  if (a >= mod) {
    a %= mod;
  }
  for (; p > 0; p /= 2) {
    if (p & 1) ret = ret * a % mod;
    a = a * a % mod;
  }
  return ret;
}
constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr auto PI = 3.14159265358979323846L;
constexpr auto oo = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr int mx = 500005;
string name[mx];
int nxt[256], n;
int dp[256];
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  for (int i = ((0)), _i = ((n)-1); i <= _i; ++i) cin >> name[i];
  int ans = 0;
  for (int c = ('a'), _c = ('z'); c <= _c; ++c) {
    for (int i = ((0)), _i = ((256) - 1); i <= _i; ++i) dp[i] = -oo;
    dp[c] = 0;
    for (int i = n - 1; i >= 0; --i) {
      char a = name[i].front(), b = name[i].back();
      int len = (int((name[i]).size()));
      dp[a] = max(dp[a], len + dp[b]);
    }
    ans = max(ans, dp[c]);
  }
  cout << ans << endl;
  return 0;
}
