Problem:
Score : 700 points

Problem Statement

Limak can repeatedly remove one of the first two characters of a string,
for example abcxyx \rightarrow acxyx \rightarrow cxyx \rightarrow cyx.

You are given N different strings S_1, S_2, \ldots, S_N.
Among N \cdot (N-1) / 2 pairs (S_i, S_j),
in how many pairs could Limak obtain one string from the other?

Constraints

2 \leq N \leq 200\,000

S_i consists of lowercase English letters a-z.

S_i \neq S_j

1 \leq |S_i|

|S_1| + |S_2| + \ldots + |S_N| \leq 10^6

Input

Input is given from Standard Input in the following format.

N
S_1
S_2
\vdots
S_N

Output

Print the number of unordered pairs (S_i, S_j)
where i \neq j and Limak can obtain one string from the other.

Sample Input 1

3
abcxyx
cyx
abc

Sample Output 1

1

The only good pair is (abcxyx, cyx).

Sample Input 2

6
b
a
abc
c
d
ab

Sample Output 2

5

There are five good pairs: (b, abc), (a, abc), (abc, c), (b, ab), (a, ab).

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define REP(i,m,n) for(int i=(int)(m); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define RREP(i,m,n) for(int i=(int)(m); i>=(int)(n); i--)
#define rrep(i,n) RREP(i,n-1,0)
#define all(v) v.begin(), v.end()
#define endk '\n'
const int inf = 1e9+7;
const ll longinf = 1LL<<60;
const ll mod = 1e9+7;
const ld eps = 1e-10;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n; cin >> n;
  vector<string> S(n);
  map<int, vector<string>> mp;
  map<int, int> size;
  rep(i, n) {
    cin >> S[i];
    reverse(all(S[i]));
    size[S[i].size()]++;
    mp[S[i].size()].push_back(S[i]);
  }
  ll ans = 0;
  rep(i, n) {
    set<char> st;
    while(S[i].size() > 0) {
      assert(S[i].size()<=2000000);
      for(char c: st) {
        string tmp = S[i]+c;
        if(size.count(tmp.size()) && find(all(mp[tmp.size()]), tmp) != mp[tmp.size()].end()) ans++;
      }
      st.insert(S[i].back());
      S[i].pop_back();
    }
    for(char c: st) {
      string tmp = S[i]+c;
      if(size.count(tmp.size()) && find(all(mp[tmp.size()]), tmp) != mp[tmp.size()].end()) ans++;
    }
    assert(st.size()<=26);
  }
  ans -= n;
  cout << ans << endk;
  return 0;
}