CF1264D1.Beautiful Bracket Sequence (easy version)

This is the easy version of this problem. The only difference is the limit of $n$ - the length of the input string. In this version, $1 \leq n \leq 2000$. The hard version of this challenge is not offered in the round for the second division.

Let's define a correct bracket sequence and its depth as follow:

• An empty string is a correct bracket sequence with depth $0$.
• If "s" is a correct bracket sequence with depth $d$ then "(s)" is a correct bracket sequence with depth $d + 1$.
• If "s" and "t" are both correct bracket sequences then their concatenation "st" is a correct bracket sequence with depth equal to the maximum depth of $s$ and $t$.

For a (not necessarily correct) bracket sequence $s$, we define its depth as the maximum depth of any correct bracket sequence induced by removing some characters from $s$ (possibly zero). For example: the bracket sequence $s = $"())(())" has depth $2$, because by removing the third character we obtain a correct bracket sequence "()(())" with depth $2$.

Given a string $a$ consists of only characters '(', ')' and '?'. Consider all (not necessarily correct) bracket sequences obtained by replacing all characters '?' in $a$ by either '(' or ')'. Calculate the sum of all the depths of all these bracket sequences. As this number can be large, find it modulo $998244353$.

Hacks in this problem in the first division can be done only if easy and hard versions of this problem was solved.

这是本题的简单版本。唯一的区别在于 $n$（输入字符串的长度）的限制。在本版本中，$1 \leq n \leq 2000$。本挑战的困难版本未在第二赛区（Div. 2）的比赛中提供。

我们定义合法括号序列及其深度如下：

• 空字符串是一个深度为 $0$ 的合法括号序列；
• 若字符串 “s” 是一个深度为 $d$ 的合法括号序列，则 “(s)” 是一个深度为 $d + 1$ 的合法括号序列；
• 若字符串 “s” 和 “t” 均为合法括号序列，则它们的拼接 “st” 也是一个合法括号序列，其深度等于 $s$ 与 $t$ 的深度的最大值。

对于一个（未必合法的）括号序列 $s$，我们定义其深度为：在 $s$ 中删去若干字符（可以为零个）后所能得到的所有合法括号序列的深度中的最大值。例如：括号序列 $s = $"())(())" 的深度为 $2$，因为删去第三个字符后可得合法括号序列 "()(())"，其深度为 $2$。

给定一个仅由字符 '(', ')' 和 '?' 组成的字符串 $a$。考虑所有将 $a$ 中每个 '?' 替换为 '(' 或 ')' 后得到的（未必合法的）括号序列。请计算所有这些括号序列的深度之和。由于结果可能很大，请对 $998244353$ 取模。

在第一赛区（Div. 1）中，本题的 Hack 仅当该选手同时通过了本题的简单版本与困难版本时才被允许。

The only line contains a non-empty string consist of only '(', ')' and '?'. The length of the string is at most $2000$.

唯一一行包含一个非空字符串，该字符串仅由 '('、')' 和 '?' 组成。字符串长度至多为 $2000$。

Print the answer modulo $998244353$ in a single line.

在一行中输出答案对 $998244353$ 取模的结果。

In the first test case, we can obtain $4$ bracket sequences by replacing all characters '?' with either '(' or ')':

• "((". Its depth is $0$;
• "))". Its depth is $0$;
• ")(". Its depth is $0$;
• "()". Its depth is $1$.

So, the answer is $1 = 0 + 0 + 0 + 1$.

In the second test case, we can obtain $4$ bracket sequences by replacing all characters '?' with either '(' or ')':

• "(((())". Its depth is $2$;
• "()()))". Its depth is $2$;
• "((()))". Its depth is $3$;
• "()(())". Its depth is $2$.

So, the answer is $9 = 2 + 2 + 3 + 2$.

在第一个测试用例中，我们可以通过将所有字符 '?' 替换为 '(' 或 ')'，得到 $4$ 个括号序列：

• "((". 其深度为 $0$；
• "))". 其深度为 $0$；
• ")(". 其深度为 $0$；
• "()". 其深度为 $1$。

因此，答案为 $1 = 0 + 0 + 0 + 1$。

在第二个测试用例中，我们可以通过将所有字符 '?' 替换为 '(' 或 ')'，得到 $4$ 个括号序列：

• "(((())". 其深度为 $2$；
• "()()))". 其深度为 $2$；
• "((()))". 其深度为 $3$；
• "()(())". 其深度为 $2$。

因此，答案为 $9 = 2 + 2 + 3 + 2$。