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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;

typedef long long ll;

int main() {
    int N;
    cin >> N;
    vector<string> S(N);
    map<string, map<char, int>> m;
    rep(i, N) {
        cin >> S[i];
        string k = S[i].substr(1, S[i].size());
        m[k][S[i][0]]++;
    }
    ll ans = 0;
    rep(i, N) {
        vector<int> c(26, 0);
        c[S[i][0] - 'a']++;
        for (int j = 1; j < S[i].size(); j++) {
            c[S[i][j] - 'a']++;
            auto m0 = m[S[i].substr(j + 1, S[i].size() - j - 1)];
            for (auto k : m0) {
                ans += 1LL * c[k.first - 'a'] * k.second;
            }
        }
    }
    cout << ans << "\n";
    /*
    for (auto i : m) {
        cout << i.first << "\n";
        for (auto j : i.second) {
            cout << j.first << " " << j.second << "\n";
        }
    }
    */
}