Bracket Subsequence
A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.You are given a regular bracket sequence $$$s$$$ and an integer number $$$k$$$. Your task is to find a regular bracket sequence of length exactly $$$k$$$ such that it is also a subsequence of $$$s$$$.It is guaranteed that such sequence always exists.
The first line contains two integers $$$n$$$ and $$$k$$$ ($$$2 \le k \le n \le 2 \cdot 10^5$$$, both $$$n$$$ and $$$k$$$ are even) — the length of $$$s$$$ and the length of the sequence you are asked to find.The second line is a string $$$s$$$ — regular bracket sequence of length $$$n$$$.
Print a single string — a regular bracket sequence of length exactly $$$k$$$ such that it is also a subsequence of $$$s$$$.It is guaranteed that such sequence always exists.
(n, k) = map(int, input().split())
print(input().replace('(', '', (n - k) // 2).replace(')', '', (n - k) // 2))
s = input().rstrip().split()
n = int(s[0])
k = int(s[1])
arr = input().rstrip()
result = []
diff = n - k
for char in arr:
    if char == '(' or diff == 0:
        result.append(char)
    else:
        result.pop()
        diff -= 2
print(''.join(result))
(n, k) = map(int, input().split())
s = input()
res = []
left = k // 2
for i in range(n):
    if left > 0 and s[i] == '(':
        left -= 1
        res.append(s[i])
    elif left == 0:
        break
    else:
        res.append(s[i])
for i in range(len(res), k):
    res.append(')')
print(''.join(res))