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 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
	static class Trie {
		public class Node {
			HashMap<Character, Node> child;
			int[] count;
			Node parent;

			public Node() {
				child = new HashMap<>();
				count = new int[26];
			}
			public Node(Node parent) {
				child = new HashMap<>();
				count = new int[26];
				this.parent = parent;
			}
		}
		
		Node root;
		public Trie() {
			root = new Node();
		}
		
		void setString(char[] c) {
			Node cur = root;
			for(int i=c.length-1; i>=0; i--) {
				if(cur.child.get(c[i])==null) {
					cur.child.put(c[i], new Node(cur));
				}
				cur = cur.child.get(c[i]);
			}
			int bit = 0;
			int index=0;
			while(cur!=null) {
				for(int i=0; i<26; i++)
					if(((1<<i)&bit)!=0)
						cur.count[i]++;
				if(index<c.length)
					bit |= 1<<(c[index]-'a');
				index++;
				cur = cur.parent;
			}
		}
		
		int getCount(char[] c) {
			Node cur = root;
			for(int i=c.length-1; i>=1; i--) {
				cur = cur.child.get(c[i]);
			}
			return cur.count[c[0]-'a'];
		}
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int N = sc.nextInt();
		char[][] s = new char[N][];
		Trie trie = new Trie();
		for(int i=0; i<N; i++) {
			s[i] = sc.next().toCharArray();
			trie.setString(s[i]);
		}
		
		long ans = 0;
		for(int i=0; i<N; i++)
			ans += trie.getCount(s[i]);
		
		System.out.println(ans - N);
		sc.close();
	}
}