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::*;
use std::collections::*;

fn run() {
    input! {
        n: usize,
        mut s: [Bytes; n],
    }
    s.iter_mut().for_each(|s| s.reverse());
    s.sort_by_key(|s| s.len());
    let mut map = BTreeMap::<usize, BTreeMap<&[u8], usize>>::new();
    let mut ans = 0;
    for s in s.iter() {
        let mut bit = 0;
        let mut pos = s.len();
        for (&len, map) in map.iter().rev() {
            while pos > len {
                pos -= 1;
                bit |= 1 << (s[pos] - b'a') as usize;
            }
            let pre = &s[..pos];
            ans += map.get(pre).map_or(0, |p| (bit & *p).count_ones());
        }
        let len = s.len();
        let pre = &s[..(len - 1)];
        let k = (s[len - 1] - b'a') as usize;
        *map.entry(len - 1).or_insert(BTreeMap::new()).entry(pre).or_insert(0) |= 1 << k;
    }
    println!("{}", ans);
}

fn main() {
    run();
}