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 OCaml (4.10.0) (judge verdict: Wrong Answer)

let () = Scanf.scanf "%d\n" @@ fun n ->
  let ss =
    Array.init n @@ fun _ ->
      Scanf.scanf "%s\n" @@ fun s ->
        Array.fold_left (fun (m, c) -> function
          | '(' -> (m, c + 1)
          | ')' -> (min m (c - 1), c - 1)) (0, 0) @@
        Array.init (String.length s) (String.get s) in
  print_endline @@
  if
    ( <> ) 0 @@
    Array.fold_right (fun (_, c) -> ( + ) c) ss 0
  then "No"
  else begin
    Array.sort (fun (m, c) (m', c') -> compare (m' + c') (m + c)) ss;
    try
      ignore
      (Array.fold_left (fun x (m, c) ->
        if x + m < 0 || x + c < 0 then raise Not_found; x + c) 0 ss);
      "Yes"
    with Not_found -> "No"
  end