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 <fstream>
#include <typeinfo>
#include <vector>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <iomanip>
#include <cctype>
#include <random>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<57;
const double pi=acos(-1);
const double eps=1e-6;
const ll mod=1e9+7;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,-0};
const int DX[8]={-1,-1,-1,0,1,1,1,0},DY[8]={1,0,-1,-1,-1,0,1,1};

class Tree{
	public:
	vector<Tree> tree;
	map<char,int> date;
	int N=0,Q=0;
	void In(Tree& t,string& s,int I){
		if(I==s.size()){
			t.Q++;
			return;
		}
		if(t.date.find(s[I])==t.date.end()){
			t.date[s[I]]=t.tree.size();
			Tree tmp;
			tmp.N=1;
			tmp.In(tmp,s,I+1);
			t.tree.push_back(tmp);
		}
		else{
			t.tree[t.date[s[I]]].N++;
			In(t.tree[t.date[s[I]]],s,I+1);
		}
	}
	int Out(Tree& t,string& S,string& s,int I){
		if(I==s.size()) return 1;
		int res=0;
		for(int i=0;i<26;i++){
			if(S[i]==s[I]) break;
			if(t.date.find(S[i])!=t.date.end()) res+=t.tree[t.date[S[i]]].N;
		}
		res+=t.Out(t.tree[t.date[s[I]]],S,s,I+1)+t.Q;
		return res;
	}
	void solve(){
		int n,q;
		vs s;
		cin>>n;
		s=vs(n);
		for(int i=0;i<n;i++){
			cin>>s[i];
			In(*this,s[i],0);
		}
		cin>>q;
		for(int i=0;i<q;i++){
			int k;
			string S;
			cin>>k>>S;
			cout<<Out(*this,S,s[k-1],0)<<endl;
		}
	}
};

int main(){
	Tree tree;
	tree.solve();
}