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: Runtime Error)

import java.util.*;
public class Main {

    /**
     * @param n int整型
     * @param m int整型
     * @param a int整型一维数组
     * @return int整型
     */
    public static int count(int n, String[] s) {
        int ret = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (convert1to2(s[i], s[j])) ret++;
            }
        }
        return ret;
    }

    public static boolean convert1to2(String a, String b) {
        int m1 = a.length(), m2 = b.length();
        if (m1 == m2) return false;
        if (b.length() == 1) {
            return a.indexOf(b) != -1;
        }

        if (a.substring(m1 - m2, m1).equals(b)) return true;
        if        (a.substring(m1 - m2 + 1, m1).equals(b.substring(1, m2))
                        && a.indexOf(b.substring(0, 1)) <= m1 - m2 && a.indexOf(b.substring(0, 1)) >= 0)
            return true;

        else
            return false;

    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String[] s = new String[n];

        for (int i = 0; i < n; i++) {
            s[0] = sc.next();
        }
        Arrays.sort(s, (s1, s2) -> s2.length() - s1.length());                
        sc.close();

        System.out.println(count(n, s));
    }
}