#include<vector>
#include<string>
#include<stdio.h>
#include<algorithm>
using namespace std;
char in[410000];
string str[110000];
struct Trie{
	int chi[26];
	int sz;
	int flag;
	Trie(){
		for(int i=0;i<26;i++)chi[i]=-1;
		sz=0;flag=0;
	}
};
Trie pool[410000];
char qs[110000][30];
pair<int,int> p[110000];
int ans[110000];
int pl[30][30];
int main(){
	int a;scanf("%d",&a);
	int ind=1;
	for(int i=0;i<a;i++){
		scanf("%s",in);
		str[i]=in;
		int at=0;
		for(int j=0;in[j];j++){
			if(pool[at].chi[in[j]-'a']==-1){
				at=pool[at].chi[in[j]-'a']=ind;
				ind++;
			}else{
				at=pool[at].chi[in[j]-'a'];
			}
			pool[at].sz++;
		}
		pool[at].flag=1;
	}
	int b;scanf("%d",&b);
	for(int i=0;i<b;i++){
		int c;scanf("%d%s",&c,qs[i]);
		c--;
		p[i]=make_pair(c,i);
	}
	std::sort(p,p+b);
	int v=0;
	for(int i=0;i<a;i++){
		for(int j=0;j<26;j++)for(int k=0;k<26;k++)pl[j][k]=0;
		int ad=0;
		int at=0;
		for(int j=0;j<str[i].size();j++){
			if(pool[at].flag)ad++;
			for(int k=0;k<26;k++){
				if(pool[at].chi[k]!=-1&&k!=str[i][j]-'a')pl[k][str[i][j]-'a']+=pool[pool[at].chi[k]].sz;
			}
			at=pool[at].chi[str[i][j]-'a'];
		}
		while(v<b&&p[v].first==i){
			int t=p[v].second;
			int tmp=ad;
			for(int j=0;j<26;j++){
				for(int k=j+1;k<26;k++){
					tmp+=pl[qs[t][j]-'a'][qs[t][k]-'a'];
				}
			}
			ans[t]=tmp;
			v++;
		}
	}
	for(int i=0;i<b;i++)printf("%d\n",ans[i]+1);
}
### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include<vector>
#include<string>
#include<stdio.h>
#include<algorithm>
using namespace std;
char in[410000];
string str[110000];
struct Trie{
	int chi[26];
	int sz;
	int flag;
	Trie(){
		for(int i=0;i<26;i++)chi[i]=-1;
		sz=0;flag=0;
	}
};
Trie pool[410000];
char qs[110000][30];
pair<int,int> p[110000];
int ans[110000];
int pl[30][30];
int main(){
	int a;scanf("%d",&a);
	int ind=1;
	for(int i=0;i<a;i++){
		scanf("%s",in);
		str[i]=in;
		int at=0;
		for(int j=0;in[j];j++){
			if(pool[at].chi[in[j]-'a']==-1){
				at=pool[at].chi[in[j]-'a']=ind;
				ind++;
			}else{
				at=pool[at].chi[in[j]-'a'];
			}
			pool[at].sz++;
		}
		pool[at].flag=1;
	}
	int b;scanf("%d",&b);
	for(int i=0;i<b;i++){
		int c;scanf("%d%s",&c,qs[i]);
		c--;
		p[i]=make_pair(c,i);
	}
	std::sort(p,p+b);
	int v=0;
	for(int i=0;i<a;i++){
		for(int j=0;j<26;j++)for(int k=0;k<26;k++)pl[j][k]=0;
		int ad=0;
		int at=0;
		for(int j=0;j<str[i].size();j++){
			if(pool[at].flag)ad++;
			for(int k=0;k<26;k++){
				if(pool[at].chi[k]!=-1&&k!=str[i][j]-'a')pl[k][str[i][j]-'a']+=pool[pool[at].chi[k]].sz;
			}
			at=pool[at].chi[str[i][j]-'a'];
		}
		while(v<b&&p[v].first==i){
			int t=p[v].second;
			int tmp=ad;
			for(int j=0;j<26;j++){
				for(int k=j+1;k<26;k++){
					tmp+=pl[qs[t][j]-'a'][qs[t][k]-'a'];
				}
			}
			ans[t]=tmp;
			v++;
		}
	}
	for(int i=0;i<b;i++)printf("%d\n",ans[i]+1);
}
```