#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <sstream>
#include <typeinfo>
#include <fstream>

#define DIV 1000000007

using namespace std;

long long N;
string S[100005];

int k[100005];
string rules[100005];

//               ch1 < ch2 なら　小さくなる
int memo[100005][26][26];

struct node {
	node *next[26] = {};
	int num[26];
	int fin;
};
node *root = new node();
 
void add(string s) {
	node *curr = root;
	for (char c : s) {
		if (curr->next[c - 'a'] == NULL) {
			curr->next[c - 'a'] = new node();
			for(int i = 0; i < 26; i++){
				curr->next[c-'a']->num[i] = 0;
			}
			curr->next[c-'a']->fin = 0;
		}
		curr->num[c-'a']++;
		curr = curr->next[c - 'a'];
	}
	curr->fin++;
}

long long count(int kdx, string rule){
	long long ret = 0;
	for(int i = 0; i < 26; i++){
		for(int j = i + 1; j < 26; j++){
			int ch1 = rule[i] - 'a';
			int ch2 = rule[j] - 'a';
			ret += memo[kdx][ch1][ch2];
		}
	}
	return ret + memo[kdx][0][0] + 1;

}


void create_table(int idx){
	node *curr = root;

	for (char c : S[idx] ) {
		long long tmp = 0;
		for(int i = 0; i < 26; i++){
			if(i != c - 'a'){
				memo[idx][i][c-'a'] += curr->num[i];
			}
		}
		memo[idx][0][0] += curr->fin;
		curr = curr->next[c - 'a'];
	}


}



int main(){
	cin >> N;

	//400000
	for(int i = 0; i < N; i++){
		cin >> S[i];
		add(S[i]);
	}

	for(int i = 0; i < N; i++){
		create_table(i);
	}


	long long Q;
	cin >> Q;
	for(int i = 0; i < Q; i++){
		cin >> k[i] >> rules[i];
		k[i]--;
	}

	/*
	for(int i = 0; i < N; i++){
		cout << "i=" << i  << endl;
		for(int j = 0; j < 26; j++){
			for(int k = 0; k < 26; k++){
				cout << "memo[" << i << "][" << (char)('a' + j) << "][" << (char)('a' + k) << "]=" << memo[i][j][k] << endl;
			}
		}
	}
	*/

	//1e5
	for(int i = 0; i < Q; i++){
		cout << count(k[i], rules[i]) << endl;
	}
}
### Prompt

Your challenge is to write 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 <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <sstream>
#include <typeinfo>
#include <fstream>

#define DIV 1000000007

using namespace std;

long long N;
string S[100005];

int k[100005];
string rules[100005];

//               ch1 < ch2 なら　小さくなる
int memo[100005][26][26];

struct node {
	node *next[26] = {};
	int num[26];
	int fin;
};
node *root = new node();
 
void add(string s) {
	node *curr = root;
	for (char c : s) {
		if (curr->next[c - 'a'] == NULL) {
			curr->next[c - 'a'] = new node();
			for(int i = 0; i < 26; i++){
				curr->next[c-'a']->num[i] = 0;
			}
			curr->next[c-'a']->fin = 0;
		}
		curr->num[c-'a']++;
		curr = curr->next[c - 'a'];
	}
	curr->fin++;
}

long long count(int kdx, string rule){
	long long ret = 0;
	for(int i = 0; i < 26; i++){
		for(int j = i + 1; j < 26; j++){
			int ch1 = rule[i] - 'a';
			int ch2 = rule[j] - 'a';
			ret += memo[kdx][ch1][ch2];
		}
	}
	return ret + memo[kdx][0][0] + 1;

}


void create_table(int idx){
	node *curr = root;

	for (char c : S[idx] ) {
		long long tmp = 0;
		for(int i = 0; i < 26; i++){
			if(i != c - 'a'){
				memo[idx][i][c-'a'] += curr->num[i];
			}
		}
		memo[idx][0][0] += curr->fin;
		curr = curr->next[c - 'a'];
	}


}



int main(){
	cin >> N;

	//400000
	for(int i = 0; i < N; i++){
		cin >> S[i];
		add(S[i]);
	}

	for(int i = 0; i < N; i++){
		create_table(i);
	}


	long long Q;
	cin >> Q;
	for(int i = 0; i < Q; i++){
		cin >> k[i] >> rules[i];
		k[i]--;
	}

	/*
	for(int i = 0; i < N; i++){
		cout << "i=" << i  << endl;
		for(int j = 0; j < 26; j++){
			for(int k = 0; k < 26; k++){
				cout << "memo[" << i << "][" << (char)('a' + j) << "][" << (char)('a' + k) << "]=" << memo[i][j][k] << endl;
			}
		}
	}
	*/

	//1e5
	for(int i = 0; i < Q; i++){
		cout << count(k[i], rules[i]) << endl;
	}
}
```