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: Time Limit Exceeded)

#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cassert>
#include <algorithm>
#include <iterator>
#include <string>
#include <vector>
#include <list>
#include <deque>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <tuple>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <complex>
#include <bitset>
#include <numeric>
using namespace std;
//emplace_back next_permutation tie get

typedef long long LL;  typedef pair<LL, LL> PLL; typedef vector<LL> VLL;
#define REP(i,n) for(int (i)=0; (i)<(n) ;++(i))
#define REPN(i,a,n) FOR((i),(a),(a)+(n))
#define FOR(i,a,b) for(int (i)=(a); (i)<(b) ;++(i))
#define PB push_back
#define MP make_pair
#define SE second
#define FI first
#define OUT(a) cout<<(a)<<endl;

unordered_map<string,int> cnt;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,q;
    string s[100010]={};
    cin >>n;
    REP(i,n){
        cin >> s[i];
        int l=s[i].length();
        REPN(j,1,l){
            cnt[s[i].substr(0,j)]++;
        }
        cnt[s[i]+'E']++;
    }

    cin >> q;
    REP(i,q){
        int k,ans=1,rank[26]={};
        string lex;
        cin >>k>>lex;
        lex="E"+lex;
        string cur=s[k-1];
        int l=cur.length();
        REP(j,26){
            rank[lex[j+1]-'a']=j+1;
        }
        REP(j,l){
            string pre=cur.substr(0,j);
            //OUT(pre)
            REP(m,rank[cur[j]-'a']){
                string t=lex.substr(m,1);
                t=pre+t;
                //OUT(t)
                ans+=cnt[t];
                //OUT(ans)
            }
        }
        cout << ans <<endl;
    }

    return 0;
}