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 Java7 (OpenJDK 1.7.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String s[] = new String[n];
		for (int i = 0; i < n; i++) {
			s[i] = sc.next();
		}
		int q = sc.nextInt();
		int k[] = new int[q];
		String p[] = new String[q];
		for (int i = 0; i < q; i++) {
			k[i] = sc.nextInt() - 1;
			p[i] = sc.next();
		}

		Tree tree = new Tree();
		for (int i = 0; i < n; i++) {
			tree.add(s[i]);
		}
		for (int i = 0; i < q; i++) {
			System.out.println(tree.order(s[k[i]], p[i]));
		}
	}
}

class Tree {
	public Node root;

	public Tree() {
		root = new Node();
	}

	public void add(String s) {
		Node node = root;
		int l = s.length();
		for (int i = 0; i <= l; i++) {
			int c;
			if (i < s.length()) {
				c = (int)(s.charAt(i) - 'a') + 1;
			} else {
				c = 0;
			}
			node.n++;
			if (node.child[c] == null) {
				node.child[c] = new Node();
			}
			node = node.child[c];
		}
		node.n++;
	}

	public int order(String s, String p) {
		int a[] = new int[27];
		a[0] = 0;
		for (int i = 0; i < 26; i++) {
			a[i + 1] = (int)(p.charAt(i) - 'a') + 1;
		}

		int o = 0;
		Node node = root;
		int l = s.length();
		for (int i = 0; i < l; i++) {
			int c = (int)(s.charAt(i) - 'a') + 1;

			for (int j = 0; a[j] != c; j++) {
				if (node.child[a[j]] != null) {
					o += node.child[a[j]].n;
				}
			}
			if (node.child[c] == null) {
				break;
			} else {
				node = node.child[c];
			}
		}
		return o + 1;
	}
}

class Node {
	public int n;
	public Node child[];

	public Node() {
		n = 0;
		child = new Node[27];
	}
}