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 {
    static int max = 100000000;
    static String[] stack = new String[max];
    static int top = 0;
    static boolean isEmpty(){return (top==0);}
    static boolean isFull(){return (top==max);}

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String[] S = new String[n];
        HashMap<String,HashSet<String>> TYPE = new HashMap<>();
        for (int i=0;i<n;i++){
            String s = sc.next();
            S[i]=s;
            push(s);
            TYPE.put(s,new HashSet<>());
            TYPE.get(s).add(s);
        }
        while (!isEmpty()){
            String ss = pop();
            if (ss.length()>1){
                String beta = ss.substring(1);
                push(beta);
                String one;
                if (ss.length()>2){
                    one = ss.charAt(0)+ss.substring(2);
                } else {
                    one = ss.substring(0,1);
                }
                push(one);
                if (!TYPE.containsKey(beta))TYPE.put(beta,new HashSet<>());
                if (!TYPE.containsKey(one))TYPE.put(one,new HashSet<>());
                TYPE.get(beta).addAll(TYPE.get(ss));
                TYPE.get(one).addAll(TYPE.get(ss));
            }
        }
        long ans = 0;
        for (int i=0;i<n;i++){
            if (TYPE.containsKey(S[i])){
                ans+=TYPE.get(S[i]).size()-1;
            }

        }
        System.out.println(ans);
    }

    static void push(String v){
        if (isFull()){
            System.out.println("error: stack is full.");
            return;
        }
        stack[top++] = v;
    }
    static String pop(){
        if (isEmpty()){
            System.out.println("error: stack is empty");
            return null;
        }
        return stack[--top];
    }

}