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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long ans = 0L;
        int n = sc.nextInt();
        PriorityQueue<String> q = new PriorityQueue<>((x,y) -> x.length()-y.length());
        for (int i = 0; i < n; i++) {
            q.add(sc.next());
        }
        Map<String, boolean[]> sufMap = new HashMap<>();
        while(!q.isEmpty()){
            String s = q.poll();
            int len = s.length();
            for (int i = 0; i < len-1; i++) {
                String suf = 0 < i ? s.substring(len-i) : "";
                if(sufMap.containsKey(suf)){
                    String res = s.substring(0, len-suf.length());
                    boolean[] b = sufMap.get(suf);
                    for (int j = 0; j < 26; j++) {
                        if(b[j] && res.contains(String.valueOf((char)('a' + j)))){
                            ans++;
                        }
                    }
                }
            }
            String suf = 1 < len ? s.substring(1) : "";
            char pre = s.charAt(0);
            if(!sufMap.containsKey(suf)) sufMap.put(suf, new boolean[26]);
            sufMap.get(suf)[pre - 'a'] = true;
        }
        System.out.println(ans);
        sc.close();
    }

}