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: Wrong Answer)

struct DFS {
    s: Vec<Vec<char>>,
}

impl DFS {
    fn new(s: Vec<Vec<char>>) -> DFS {
        DFS { s }
    }

    fn search(&mut self, d: usize, ids: &Vec<usize>, mut vb: Vec<bool>) -> i64 {
        let mut res = 0;

        if d > 0 {
            let k = self.s[ids[0]][d - 1] as usize - 'a' as usize;
            vb[k] = false;
        }

        let mut next = vec![vec![]; 26];
        for &i in ids {
            if self.s[i].len() == d {
                res += ids.len() as i64 - 1;
                continue;
            }
            let k = self.s[i][d] as usize - 'a' as usize;
            next[k].push(i);
        }

        for i in 0..26 {
            if vb[i] {
                res += next[i].len() as i64;
            }
        }

        for &i in ids {
            if self.s[i].len() == d + 1 {
                let k = self.s[i][d] as usize - 'a' as usize;
                vb[k] = true;
            }
        }

        for i in 0..26 {
            if !next[i].is_empty() {
                res += self.search(d + 1, &next[i], vb.clone());
            }
        }

        res
    }
}

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 dfs = DFS::new(s);
    let ans = dfs.search(0, &(0..n).collect(), vec![false; 26]);
    println!("{}", ans);
}