codeforces B. Petya and Exam （模拟）

B. Petya and Exam

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one..

There is a glob pattern in the statements (a string consisting of lowercase English letters, characters "?" and "*"). It is known that character "*" occurs no more than once in the pattern.

Also, n query strings are given, it is required to determine for each of them if the pattern matches it or not.

Everything seemed easy to Petya, but then he discovered that the special pattern characters differ from their usual meaning.

A pattern matches a string if it is possible to replace each character "?" with one good lowercase English letter, and the character "*" (if there is one) with any, including empty, string of bad lowercase English letters, so that the resulting string is the same as the given string.

The good letters are given to Petya. All the others are bad.

Input

The first line contains a string with length from 1 to 26 consisting of distinct lowercase English letters. These letters are good letters, all the others are bad.

The second line contains the pattern — a string s of lowercase English letters, characters "?" and "*" (1 ≤ |s| ≤ 105). It is guaranteed that character "*" occurs in s no more than once.

The third line contains integer n (1 ≤ n ≤ 105) — the number of query strings.

n lines follow, each of them contains single non-empty string consisting of lowercase English letters — a query string.

It is guaranteed that the total length of all query strings is not greater than 105.

Output

Print n lines: in the i-th of them print "YES" if the pattern matches the i-th query string, and "NO" otherwise.

You can choose the case (lower or upper) for each letter arbitrary.

Examples

input

ab
a?a
2
aaa
aab

output

YES
NO

input

abc
a?a?a*
4
abacaba
abaca
apapa
aaaaax

output

NO
YES
NO
YES

Note

In the first example we can replace "?" with good letters "a" and "b", so we can see that the answer for the first query is "YES", and the answer for the second query is "NO", because we can't match the third letter.

Explanation of the second example.

• The first query: "NO", because character "*" can be replaced with a string of bad letters only, but the only way to match the query string is to replace it with the string "ba", in which both letters are good.
• The second query: "YES", because characters "?" can be replaced with corresponding good letters, and character "*" can be replaced with empty string, and the strings will coincide.
• The third query: "NO", because characters "?" can't be replaced with bad letters.

• The fourth query: "YES", because characters "?" can be replaced with good letters "a", and character "*" can be replaced with a string of bad letters "x".

题目链接：http://codeforces.com/contest/832/problem/B

题目大意：第一行代表good word，第二行是模式串，?可以代替任意一个good word，*可以匹配为空，也可以代替任意一个全部都是bad word的子串，给定n个主串，问能否和模式串匹配，能就输出“YES”，不能就输出“NO”。

解题思路：模拟。难点在于＊处的判断，考虑＊匹配为空和为bad word的子串两种情况，注意除了＊匹配的部分外，其他部分的字符是一一对应的，即其他部分的长度相等，可以从这一点入手。

代码如下：

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
using namespace std;
int a[26];
int main(void){
	string gd,s;
	cin >> gd;
	memset(a,0,sizeof(a));
	int len;
	len = gd.length();
	bool flog = false;
	for(int i=0;i<len;i++){   //标记good word
		a[gd[i]-'a'] = 1;
	}
	cin >> s;
	len = s.length();
	for(int i=0;i<len;i++){    //判断模式串中是否有＊
		if(s[i]=='*'){
			flog = true;
			break;
		}
	} 
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		string cur;
		cin >> cur;
		bool p = true;
		int ls = s.length();
		int lc = cur.length();
		if(ls > lc+1 || ls == lc+1 && !flog ){  //如果模式串中没有＊，串长相等才能匹配，如果有＊，模式串串长要大于主串串长减1才匹配
			printf("NO\n");
			continue;
		}
		int k = 0;
		for(int j=0;j<ls,k<lc;j++,k++){
			if(s[j]<='z' && s[j] >= 'a'){   //主串的字母要和模式串的字母匹配
				if(s[j] != cur[k]){
					p = false;
					break;
				}
			}
			else if(s[j] == '?'){        //?匹配good word
				int x = cur[k]-'a';
				if(!a[x]){
					p = false;
					break;
				}
			}
			else if(s[j] == '*'){      
				int x = cur[k]-'a';
				for( ;k < lc-(ls-j)+1 ; k++){ //＊后面剩下的长度要相等才匹配
					x = cur[k]-'a';
					if(a[x]){          //＊匹配bad word
						p = false;
						break;
					}
				}
				if(p){
					if(lc - k < ls - j )   //＊匹配为空的情况
						k--;
					continue;
				}
			}
			if(j == ls-1 && k < lc-1){
				p = false;
				break;
			}
			if(k == lc-1 && (j < ls-2 || j == ls-2 && s[ls-1]!='*')){
				p = false;
				break;
			}
 		}
		if(p)
			printf("YES\n");
		else
			printf("NO\n");
	}
}