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: Accepted)

use proconio::marker::*;
use proconio::*;

type Link = Option<Box<Node>>;

const F: usize = 26;

#[derive(Default)]
struct Node {
    cnt: [u32; F],
    next: [Link; F],
}

impl Node {
    fn insert(&mut self, s: &[u8]) -> (usize, usize) {
        if s.is_empty() {
            return (0, 0);
        }
        let k = (s[0] - b'a') as usize;
        let (mut bit, mut ans) = self.next[k]
            .get_or_insert_with(|| Box::new(Node::default()))
            .insert(&s[1..]);
        if s.len() == 1 {
            ans += self.cnt[k] as usize;
        }
        bit |= 1 << k;
        for (i, c) in self.cnt.iter_mut().enumerate() {
            *c += (bit >> i & 1) as u32;
        }
        (bit, ans)
    }
}

fn run() {
    input! {
        n: usize,
        mut s: [Bytes; n],
    }
    let mut root = Box::new(Node::default());
    s.sort_by_key(|s| s.len());
    let mut ans = 0;
    for s in s.iter_mut().rev() {
        s.reverse();
        ans += root.insert(s).1;
    }
    println!("{}", ans);
}

fn main() {
    let stack_size = (1 << 28) + (1 << 27);
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| run()).unwrap().join().unwrap();
}