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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

// -*- coding:utf-8-unix -*-

use proconio::{fastout, input};
use std::collections::{BTreeSet, HashSet};

#[fastout]
fn main() {
    input! {
        n: usize,
        s_str: [String; n],
    }
    let mut s: Vec<String> = Vec::new();
    for si in s_str {
        s.push(si.chars().rev().collect());
    }
    s.sort_unstable_by_key(|x| x.len());
    let mut ans: u64 = 0;
    let mut already_seen: BTreeSet<&str> = BTreeSet::new();
    for i in 1..=n {
        let mut nowstr: String = s[i - 1].clone();
        let nownagasa = nowstr.len();
        let mut have_in_prefix: Vec<bool> = vec![false; 26];
        let chars_table = vec![
            'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q',
            'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
        ];
        have_in_prefix[(nowstr.pop().unwrap() as usize) - ('a' as usize)] = true;
        for _j in 1..nownagasa {
            have_in_prefix[(nowstr.pop().unwrap() as usize) - ('a' as usize)] = true;
            for k in 0..26 {
                if !have_in_prefix[k as usize] {
                    continue;
                }
                nowstr.push(chars_table[k]);
                if already_seen.contains(nowstr.as_str()) {
                    ans += 1;
                }
                nowstr.pop();
            }
        }
        already_seen.insert(s[i - 1].as_str());
    }
    println!("{}", ans);
}