Problem:
Score : 400 points

Problem Statement

You are given a string S of length N consisting of ( and ). Your task is to insert some number of ( and ) into S to obtain a correct bracket sequence.

Here, a correct bracket sequence is defined as follows:

() is a correct bracket sequence.

If X is a correct bracket sequence, the concatenation of (, X and ) in this order is also a correct bracket sequence.

If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence.

Every correct bracket sequence can be derived from the rules above.

Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

Constraints

The length of S is N.

1 ≤ N ≤ 100

S consists of ( and ).

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the lexicographically smallest string among the shortest correct bracket sequences that can be obtained by inserting some number of ( and ) into S.

Sample Input 1

3
())

Sample Output 1

(())

Sample Input 2

6
)))())

Sample Output 2

(((()))())

Sample Input 3

8
))))((((

Sample Output 3

(((())))(((())))

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n = gets.chomp.to_i
brackets = gets.chomp
partical_brackets = brackets

def next_short_brackets(brackets)
  return brackets.split('()').join
end

while partical_brackets.split('()').count != 1 && !partical_brackets.empty? do
  partical_brackets = next_short_brackets(partical_brackets)
end

if partical_brackets == '()'
  partical_brackets = ''
end

print partical_brackets

array = partical_brackets.split('')
lefthand_counts = array.count {|item| item == ')'}
righthand_counts = array.count {|item| item == '('}

input = brackets.split('')
lefthand_insertion = ['('] * lefthand_counts
righthand_insertion = [')'] * righthand_counts

answer = lefthand_insertion.concat(input).concat(righthand_insertion).join
print answer