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

type Link = Option<Box<Node>>;

const F: usize = 26;

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

fn run() {
    input! {
        n: usize,
        mut s: [Bytes; n],
    }
    s.sort_by_key(|s| s.len());
    let mut root = Some(Box::new(Node::default()));
    let mut ans = 0;
    for s in s.iter_mut().rev() {
        s.reverse();
        let mut now = root.take().unwrap();
        let mut stack = vec![];
        for c in s.iter() {
            let k = (*c - b'a') as usize;
            if now.next[k].is_none() {
                now.next[k] = Some(Box::new(Node::default()));
            }
            let next = now.next[k].take().unwrap();
            stack.push(now);
            now = next;
        }
        let mut bit = 0;
        for (i, c) in s.iter().rev().enumerate() {
            let mut parent = stack.pop().unwrap();
            let k = (*c - b'a') as usize;
            if i == 0 {
                ans += parent.cnt[k] as usize;
            }
            bit |= 1 << k;
            for (j, cnt) in parent.cnt.iter_mut().enumerate() {
                *cnt += ((bit >> j) & 1) as u32;
            }
            parent.next[k] = Some(now);
            now = parent;
        }
        root = Some(now);
    }
    println!("{}", ans);
}

fn main() {
    run();
}