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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

from collections import defaultdict

"""
・trie木
・各ノードから、アルファベットごとの子ノードの番号
・親ノードの番号
"""

N = int(readline())
data = tuple(read().split())
S = data[:N]
Q = data[N]
it = iter(data[N+1:])
query = zip(it,it)

class Node():
    def __init__(self,parent):
        self.parent=parent
        self.child = defaultdict(int)
        self.is_word_node = False
        self.Nword = 0 # 部分木にあるword_nodeの個数
        
    def __repr__(self):
        return 'parent:{}\nchild:{}\nNword:{}'.format(self.parent,self.child,self.Nword)

def add(word):
    n = 0
    next_idx = len(nodes)
    for x in word:
        nodes[n].Nword += 1
        c = nodes[n].child[x]
        if c == 0:
            c = next_idx
            nodes[n].child[x] = c
            nodes.append(Node(n))
            next_idx += 1
        n = c
    nodes[n].is_word_node = True
    nodes[n].Nword += 1

def solve(word):
    """
    ・wordに沿って進む。
    ・「char1<char2ならば自身より弱い文字列1個」という情報を調べる・
    ・自身のprefixの個数も調べる
    """
    data = defaultdict(int)
    prefix = 0
    n = 0
    for x in word:
        # 進む前に、他のルートの文字列数も確認する
        for k,v in nodes[n].child.items():
            if k == x:
                continue
            data[1000*k+x] += nodes[v].Nword # k<xのときの加算量
        # prefixを見つけたら加算
        if nodes[n].is_word_node:
            prefix += 1
        # 進む
        n = nodes[n].child[x]
    return data, prefix

root = Node(0)
nodes = [root]

for word in S:
    add(word)

compare_data = []
prefix_cnt = []
for word in S:
    data,pref = solve(word)
    compare_data.append(data)
    prefix_cnt.append(pref)