CF1023A.Single Wildcard Pattern Matching

You are given two strings $s$ and $t$. The string $s$ consists of lowercase Latin letters and at most one wildcard character '*', the string $t$ consists only of lowercase Latin letters. The length of the string $s$ equals $n$, the length of the string $t$ equals $m$.

The wildcard character '*' in the string $s$ (if any) can be replaced with an arbitrary sequence (possibly empty) of lowercase Latin letters. No other character of $s$ can be replaced with anything. If it is possible to replace a wildcard character '*' in $s$ to obtain a string $t$, then the string $t$ matches the pattern $s$.

For example, if $s=$"aba*aba" then the following strings match it "abaaba", "abacaba" and "abazzzaba", but the following strings do not match: "ababa", "abcaaba", "codeforces", "aba1aba", "aba?aba".

If the given string $t$ matches the given string $s$, print "YES", otherwise print "NO".

给你两个字符串 $s$ 和 $t$。字符串 $s$ 由小写拉丁字母组成，且最多包含一个通配符字符 '*'；字符串 $t$ 仅由小写拉丁字母组成。字符串 $s$ 的长度为 $n$，字符串 $t$ 的长度为 $m$。

字符串 $s$ 中的通配符 '*'（如果存在）可以被替换为任意长度（可能为空）的小写拉丁字母序列。$s$ 中的其他字符均不可被替换。若能将 $s$ 中的通配符 '*' 替换为某个字符串，使得结果恰好等于 $t$，则称字符串 $t$ 与模式 $s$ 匹配。

例如，若 $s=$"aba*aba"，则以下字符串与之匹配："abaaba"、"abacaba" 和 "abazzzaba"；但以下字符串不匹配："ababa"、"abcaaba"、"codeforces"、"aba1aba"、"aba?aba"。

若给定的字符串 $t$ 与给定的字符串 $s$ 匹配，则输出 "YES"；否则输出 "NO"。

The first line contains two integers $n$ and $m$ ($1 \le n, m \le 2 \cdot 10^5$) — the length of the string $s$ and the length of the string $t$, respectively.

The second line contains string $s$ of length $n$, which consists of lowercase Latin letters and at most one wildcard character '*'.

The third line contains string $t$ of length $m$, which consists only of lowercase Latin letters.

第一行包含两个整数 $n$ 和 $m$（$1 \le n, m \le 2 \cdot 10^5$），分别表示字符串 $s$ 和字符串 $t$ 的长度。

第二行包含一个长度为 $n$ 的字符串 $s$，由小写拉丁字母组成，且最多包含一个通配符字符 '*'。

第三行包含一个长度为 $m$ 的字符串 $t$，仅由小写拉丁字母组成。

Print "YES" (without quotes), if you can obtain the string $t$ from the string $s$. Otherwise print "NO" (without quotes).

如果可以从字符串 $s$ 得到字符串 $t$，则输出 "YES"（不带引号）；否则输出 "NO"（不带引号）。

In the first example a wildcard character '*' can be replaced with a string "force". So the string $s$ after this replacement is "codeforces" and the answer is "YES".

In the second example a wildcard character '*' can be replaced with an empty string. So the string $s$ after this replacement is "vkcup" and the answer is "YES".

There is no wildcard character '*' in the third example and the strings "v" and "k" are different so the answer is "NO".

In the fourth example there is no such replacement of a wildcard character '*' that you can obtain the string $t$ so the answer is "NO".

在第一个例子中，通配符字符 '*' 可以被替换为字符串 "force"。因此，替换后的字符串 $s$ 为 "codeforces"，答案为 "YES"。

在第二个例子中，通配符字符 '*' 可以被替换为空字符串。因此，替换后的字符串 $s$ 为 "vkcup"，答案为 "YES"。

在第三个例子中，字符串中没有通配符字符 '*'，且字符串 "v" 与 "k" 不同，因此答案为 "NO"。

在第四个例子中，不存在对通配符字符 '*' 的任何替换方式，使得能得到字符串 $t$，因此答案为 "NO"。