Problem:
Score : 1100 points

Problem Statement

There are N strings of lowercase alphabet only. The i-th string is S_i. Every string is unique.

Provide answers for the Q queries below. The i-th query has the following format:

Query: An integer k_i and a string p_{i,1}p_{i,2}...p_{i,26} that results from permuting {a,b,...,z} are given. Output the sequence of the string S_{k_i} among the N strings in lexicographical order when the literal sequence is p_{i,1}<p_{i,2}<...<p_{i,26}.

Constraints

1 ≦ N,Q ≦ 100000

1 ≦ |S_i| (1 ≦ i ≦ N)

S_i (1 ≦ i ≦ N) is a string of lowercase alphabet.

The sum of |S_i| is no more than 400000.

Every S_i is unique.

1 ≦ k_i ≦ N (1 ≦ i ≦ Q)

For all 1 ≦ i ≦ Q, p_{i,1}p_{i,2}...p_{i,26} is a permutation of abcd...z.

Input

Inputs are provided from standard inputs in the following form.

N
S_1
:
S_N
Q
k_1 p_{1,1}p_{1,2}...p_{1,26}
:
k_Q p_{Q,1}p_{Q,2}...p_{Q,26}

Output

Output Q lines.

On line i, for the i-th query, output an integer indicating the sequence of the string S_{k_i} among the N strings in lexicographical order.

Sample Input 1

5
aa
abbaa
abbba
aaab
aaaaaba
5
1 abcdefghijklmnopqrstuvwxyz
2 bacdefghijklmnopqrstuvwxyz
3 abcdefghijklmnopqrstuvwxyz
4 bacdefghijklmnopqrstuvwxyz
5 abcdefghijklmnopqrstuvwxyz

Sample Output 1

1
2
5
4
2

When the literal sequence is a < b, sorting the input strings in lexicographical order yields aa, aaaaaba, aaab, abbaa, abbba. The answers to queries 1, 3, and 5 are thus 1, 5, and 2, respectively.

When the literal sequence is b < a, sorting the input strings in lexicographical order yields abbba, abbaa, aa, aaab, aaaaaba. The answers to queries 2 and 4 are thus 2 and 4, respectively.

Sample Input 2

8
abrakatabra
abadaba
abracadabra
atcoder
grand
contest
ababa
a
6
3 abcdefghijklmnopqrstuvwxyz
6 qwertyuiopasdfghjklzxcvbnm
8 poiuytrewqlkjhgfdsamnbvcxz
2 qazwsxedcrfvtgbyhnujmikolp
1 plokmijnuhbygvtfcrdxeszwaq
4 mnbvcxzasdfghjklpoiuytrewq

Sample Output 2

4
8
2
3
4
7

Solution in Rust (1.42.0) (judge verdict: Accepted)

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

type Ref = Box<Node>;
type Link = Option<Ref>;

const F: usize = 26;

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

impl Node {
    fn new() -> Ref {
        Box::new(Node::default())
    }
    fn insert(&mut self, s: &[u8]) {
        let mut po = self;
        for c in s.iter() {
            po.cnt += 1;
            let k = (*c - b'a') as usize;
            po = po.next[k].get_or_insert_with(Node::new);
        }
        po.cnt += 1;
    }
    fn dfs(&self, s: &[u8]) -> (u32, [[u32; F]; F]) {
        let mut add = 0;
        let mut cnt = [[0u32; F]; F];
        let mut po = self;
        for c in s.iter() {
            let k = (*c - b'a') as usize;
            add += po.cnt - po.next.iter().map(|p| p.as_ref().map_or(0, |p| p.cnt)).sum::<u32>();
            for (x, (cnt, next)) in cnt[k].iter_mut().zip(po.next.iter()).enumerate() {
                if k == x {
                    continue;
                }
                *cnt += next.as_ref().map_or(0, |p| p.cnt);
            }
            po = po.next[k].as_ref().unwrap();
        }
        (add, cnt)
    }
}

#[fastout]
fn run() {
    input! {
        n: usize,
        s: [Bytes; n],
        q: usize,
        ask: [(Usize1, Bytes); q],
    }
    let mut root = Node::new();
    for s in s.iter() {
        root.insert(s);
    }
    let root = root;
    let mut memo = vec![];
    for s in s.iter() {
        memo.push(root.dfs(s));
    }
    for (k, p) in ask {
        let memo = &memo[k];
        let mut ans = memo.0 + 1;
        for (i, x) in p.iter().enumerate() {
            let x = (*x - b'a') as usize;
            let cnt = &memo.1[x];
            for y in p.iter().take(i) {
                let y = (*y - b'a') as usize;
                ans += cnt[y];
            }
        }
        println!("{}", ans);
    }
}

fn main() {
    run();
}