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.

Examples

Input

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


Output

1
2
5
4
2


Input

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


Output

4
8
2
3
4
7
# 文字列の問題で、同じような文字列をまとめたいので、
# どう考えても Trie を使いたい気持ちになる

# あとは適当にやったらなんとかなった
# 1100 点の割には簡単っぽい

class TrieNode:
    def __init__(self):
        self.child = {}
        self.count = 0  # 自身を含めた部分木の大きさ

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def add(self, s):
        node = self.root
        node.count += 1
        for c in s:
            c = ord(c) - 97
            if c not in node.child:
                node.child[c] = TrieNode()
            node = node.child[c]
            node.count += 1
        node.child[26] = TrieNode()
        node.child[26].count += 1

    def query(self, s):
        res = [[0]*27 for _ in range(26)]
        node = self.root
        for c in s:
            c = ord(c) - 97
            res_c = res[c]
            for i, child in node.child.items():
                res_c[i] += child.count
            node = node.child[c]
        return res

from itertools import groupby
import sys

def main():
    input = sys.stdin.readline
    N = int(input())
    S = [input()[:-1] for _ in range(N)]
    trie = Trie()
    for s in S:
        trie.add(s)

    Q = int(input())
    KP = [input().split() for _ in range(Q)]
    Ans = [-1] * Q
    Idx_Q = sorted(range(Q), key=lambda x: KP[x][0])
    for v, g in groupby(Idx_Q, key=lambda x: KP[x][0]):
        K = int(v) - 1
        L_K = trie.query(S[K])
        for idx in g:
            _, P = KP[idx]
            P = [ord(c) - 97 for c in P]
            ans = 1
            for i, pi in enumerate(P):
                for pj in P[:i]:
                    ans += L_K[pi][pj]
                ans += L_K[pi][26]
            Ans[idx] = ans
    print("\n".join(map(str, Ans)))

main()
