Regular Bracket Sequence
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".You are given a sequence $$$s$$$ of $$$n$$$ characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.Determine if it is possible to obtain an RBS after these replacements.
In the first test case, the sequence is already an RBS.In the third test case, you can obtain an RBS as follows: ()() or (()).In the fourth test case, you can obtain an RBS as follows: ()().
The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.Each test case consists of one line containing $$$s$$$ ($$$2 \le |s| \le 100$$$) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.
For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
t = int(input())
while t > 0:
    s = input()
    if len(s) % 2 == 0 and s[0] != ')' and (s[len(s) - 1] != '('):
        print('YES')
    else:
        print('NO')
    t -= 1
t = int(input())
while t > 0:
    t -= 1
    s = input()
    if len(s) % 2 == 1:
        print('NO')
    elif s[0] != ')' and s[-1] != '(':
        print('YES')
    else:
        print('NO')
for _ in range(int(input())):
    s = input()
    if len(s) % 2 != 1 and s[0] != ')' and (s[-1] != '('):
        print('Yes')
        continue
    print('No')