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)

line = readlines.map(&:chomp)

N = line[0].to_i

num_open = []
num_close = []
for i in 1..N do
  open = 0
  close = 0
  if line[i] == nil then
    len = 0
  else
    len = line[i].length
  end
  for j in 1..len do
    if line[i][j-1] == ")" and open == 0 then
      close += 1
    elsif line[i][j-1] == ")" then
      open -= 1
    else
      open += 1
    end
  end
  num_open.push(open)
  num_close.push(close)
end

rest_open = 0
rest_close = 0
for i in 1..N do
  last_open = rest_open
  last_close = rest_close
  if last_open >= num_close[i-1] then
    rest_open -= num_close[i-1]
  else
    rest_open = 0
    rest_close += num_close[i-1] - rest_open
  end
  if last_close >= num_open[i-1] then
    rest_close -= num_open[i-1]
  else
    rest_close = 0
    rest_open += num_open[i-1] - rest_close
  end
end
    
if rest_open == 0 and rest_close == 0 then
  print "Yes"
else
  print "No"
end