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 = gets.to_i

up = []
stay = []
down = []

def sort_rising(arr)
  arr.sort_by! { |a| [-a[0], -a[1]]}
end

def sort_falling(arr)
  arr.sort_by! { |a| [a[0], -a[1]]}
end

n.times do |i|
  mx = 0
  mn = 0
  cur = 0

  s = gets.rstrip
  s.length.times do |i|
    c = s[i]
    if c == '('
      cur += 1
    else # == ')'
      cur -= 1
    end

    mx = [mx, cur].max
    mn = [mn, cur].min
  end

  ore = [mn, cur]

  up   << ore if cur > 0
  stay << ore if cur == 0
  down << ore if cur < 0
end

sim = sort_rising(up) + stay + sort_falling(down)

# p [up, stay, down]


depth = 0
sim.each do |a|
  mn, cur = a

  if depth + mn < 0
    puts 'No'
    exit
  end

  depth += cur
  if depth < 0 
    puts 'No'
    exit
  end
end

if depth == 0
  puts 'Yes'
else
  puts 'No'
end