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)

fn main() {
    let n: usize = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().parse().unwrap()
    };

    let mut s: Vec<Vec<char>> = Vec::with_capacity(n);
    for _ in 0..n {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        s.push(buf.trim_end().chars().rev().collect());
    }

    let mut g = vec![vec![0; 26]];
    let mut v = vec![vec![]; n];
    let mut leaves = std::collections::HashSet::new();
    for i in 0..n {
        let mut now = 0;
        for &c in &s[i] {
            v[i].push(now);
            let k = c as usize - 'a' as usize;
            if g[now][k] == 0 {
                g[now][k] = g.len();
                g.push(vec![0; 26]);
            }
            now = g[now][k];
        }
        leaves.insert(now);
    }

    let mut ans = 0;
    for i in 0..n {
        let mut check = vec![false; 26];
        for j in (0..s[i].len()).rev() {
            check[s[i][j] as usize - 'a' as usize] = true;
            let now = v[i][j];
            for k in 0..26 {
                if check[k] && leaves.contains(&g[now][k]) {
                    ans += 1;
                }
            }
        }
    }
    println!("{}", ans - n);
}