Petya and Exam(Codeforces Round #425 (Div. 2))

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 onegood lowercase English letter, and the character "*" (if there is one) with any, including empty, string ofbad 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 to26 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 ins 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 than10⁵.

Output

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

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

Examples

Input

aba?a2aaaaab

Output

YESNO

Input

abca?a?a*4abacabaabacaapapaaaaaax

Output

NOYESNOYES

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".

题意理解：先给一串字符 A，A字符串里的字符都是good的字母，其他的字母就都是bad的字母，第二行给定一个字符串B，B中的 ？可以替换成 good的字母，*可以是空的或者是一个字符串（不一定是单个字符），再输入测试，测试的字符串C一定要和变化后的字符串B的长度相同。

解题思路： 就只注意* 的变化。还有*可以替换成字符串。

#include<stdio.h>#include<string.h>char a[30];char b[100010];char c[100010];int book[200];int main(){   int now=0;   scanf("%s",a);   int n1=strlen(a);   for(int i=0;i<n1;i++)   {       book[a[i]]=1;   }   scanf("%s",b);   int n2=strlen(b);   int f=0;   for(int i=0;i<n2;i++)   {       if(b[i]=='*')       {           f=1;           break;       }   }   int flag=0;   int t;   scanf("%d",&t);   while(t--)   {       flag=0;       now=0;       memset(c,0,sizeof(c));       scanf("%s",c);       int n3=strlen(c);       if(f==0&&n2!=n3)       {           printf("NO\n");           continue;       }       if(f==1&&n2-n3>1)       {           printf("NO\n");           continue;       }       for(int i=0;i<n2;i++)       {           if(b[i]=='?')           {               if(!(book[c[i+now]]))               {                   flag=1;                   break;               }           }           else if(b[i]=='*')           {               int temp=n3-n2+1;               now=temp-1;               for(int j=i;j<(i+temp);j++)               {                   if(book[c[j]]==1)                   {                       flag=1;                       break;                   }               }           }           else           {               if(b[i]!=c[i+now])               {                   flag=1;                   break;               }           }           if(flag==1)            break;       }       if(flag==1)       {           printf("NO\n");       }       else       {           printf("YES\n");       }   }}