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)

num_strings = gets.to_i

strings = num_strings.times.map do |i|
  gets.chomp
end

left = Array.new(num_strings) { 0 }
right = Array.new(num_strings) { 0 }

strings.each_with_index do |s, i|
  r = 0
  l = 0
  s.length.times do |j|
    case s[j]
    when "("
      r += 1
    when ")"
      if r == 0
        l += 1
      else
        r -= 1
      end
    end
  end

  left[i] = l
  right[i] = r
end

ary = []

num_strings.times do |i|
  ary.push([left[i], right[i]])
end

ary.sort! do |a, b|
  b[1] - b[0] <=> a[1] - a[0]
end

rem = 0

res = true

num_strings.times do |i|
  taken = nil
  ary.each_with_index do |e, i|
    if rem >= e[0]
      rem -= e[0]
      rem += e[1]
      taken = i
      break
    end
  end

  unless taken
    res = false
    break
  end

  ary.delete_at(taken)
end

if res && rem == 0
  puts "Yes"
else
  puts "No"
end