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: Time Limit Exceeded)

import sys
readline = sys.stdin.readline
class Node:
    def __init__(self, sigma, depth):
        self.end = False
        self.child = [None] * sigma
        self.depth = depth
        self.count = 0
    def __setitem__(self, i, x):
        self.child[i] = x
    def __getitem__(self, i):
        return self.child[i]

class Trie():
    def __init__(self, sigma):
        self.sigma = sigma
        self.root = Node(sigma, 0)
    
    def add(self, S):
        vn = self.root
        vn.count += 1
        for cs in S:
            if vn[cs] is None:
                vn[cs] = Node(self.sigma, vn.depth + 1)
            vn = vn[cs]
            vn.count += 1
        vn.end = True

T = Trie(26)
N = int(readline())
Di = [None]*N

for i in range(N):
    S = list(map(lambda x: ord(x)-97, readline().strip()))
    Di[i] = S
    T.add(S)

Q = int(readline())
Ans = [None]*Q
Query = [[] for _ in range(N)]
for qu in range(Q):
    k, p = readline().strip().split()
    k = int(k)-1
    p = list(map(lambda x:ord(x)-97, p))
    pinv = [None]*26
    for i in range(26):
        pinv[p[i]] = i
    Query[k].append((qu, pinv))

for i in range(N):
    if not Query[i]:
        continue
    L, P = map(list, zip(*Query[i]))
    M = len(P)
    vn = T.root
    ans = [0]*M
    for si in Di[i]:        
        for idx in range(M):
            pinv = P[idx]
            for al in range(26):
                if vn[al] and pinv[al] < pinv[si]:
                    ans[idx] += vn[al].count
        vn = vn[si]
        if vn.end:
            ans = [1+a for a in ans]
    for j in range(M):
        Ans[L[j]] = ans[j]
print('\n'.join(map(str, Ans)))