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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define rep(i,x,y) for(int i=(x);i<(y);++i)
#define debug(x) #x << "=" << (x)

#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif

const int inf=1e9;
const int64_t inf64=1e18;
const double eps=1e-9;

template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){
    os << "[";
    for (const auto &v : vec) {
    	os << v << ",";
    }
    os << "]";
    return os;
}

class trie{
public:
    trie* next[26];
    int size;

    trie(){
        fill(next,next+26,(trie*)0);
        size=0;
    }
    void insert(const string &str){
        insert(str,0);
    }
    bool find(const string &str,const string &dic){
        return find(str,0,dic);
    }
    void insert(const string &str,const int idx){
        if(idx==str.size()){
            ++size;
            return;
        }
        if(this->next[str[idx]-'a']==NULL) this->next[str[idx]-'a']=new trie();
        ++size;
        this->next[str[idx]-'a']->insert(str,idx+1);
    }
    int find(const string &str,const int idx,const string &dic){
        if(idx==str.size()) return 0;
        if(this->next[str[idx]-'a']==NULL) return 0;
        int sum=0;
        rep(i,0,26){
            if(dic[i]==str[idx]) break;
            if(this->next[dic[i]-'a']!=NULL) sum+=this->next[dic[i]-'a']->size;
        }
        return size+this->next[str[idx]-'a']->find(str,idx+1,dic);
    }
};

void solve(){
    int n;
    cin >> n;
    vector<string> s(n);
    trie t;
    rep(i,0,n){
        cin >> s[i];
        t.insert(s[i]);
    }
    int q;
    cin >> q;
    rep(i,0,q){
        int k;
        string q;
        cin >> k >> q;
        --k;
        cout << t.find(s[k],q) << endl;
    }
}

int main(){
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(10);
    solve();
    return 0;
}