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>
#define rep(i,n) for(int i=0;i<(int)n;i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    int n;
    cin >> n;
    vector<string>s(n);
    rep(i,n) cin >> s[i];
    ll ans = 0;
    rep(i,26) {
        char now_let = 'a' + i;
        map<string, int>bun;
        rep(j, n) {
            int cnt = 1e9;
            rep(k, s[j].size()) {
                if (now_let == s[j][k]) {
                    cnt = k;
                    break;
                }
            }
            string now = "";
            for (int k = s[j].size() - 1; k >= cnt; k--) {
                bun[now]++;
                now += s[j][k];
            }
        }
        rep(j, n) {
            if (now_let == s[j][0]) {
                if (s[j].size() == 1) ans += (ll)max(0, bun[""] - 1);
                else {
                    string tansaku = s[j].substr(1, s[j].size() - 1);
                    reverse(tansaku.begin(), tansaku.end());
                    ans += (ll)max(0, bun[tansaku] - 1);
                }
            }
        }
    }
    cout << ans << endl;
    return 0;
}