Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

N = $stdin.gets.to_i
total_close_count = 0
total_open_count = 0
min_close_count = nil
1.upto(N) do
  bs = $stdin.gets.strip
  close_count = 0
  open_count = 0
  bs.each_char do |b|
    case b
    when '('
      open_count += 1
    when ')'
      if open_count > 0
        open_count -= 1
      else
        close_count += 1
      end
    end
  end
  if close_count == 0 && open_count == 0
    next
  elsif !min_close_count || min_close_count > close_count
    min_close_count = close_count
  end
  total_close_count += close_count
  total_open_count += open_count
end
if total_open_count == total_close_count
  if total_open_count == 0 || min_close_count == 0
    puts 'Yes'
  else
    puts 'No'
  end
else
  puts 'No'
end