output_description: Print the lexicographically smallest string among the shortest correct bracket
  sequences that can be obtained by inserting some number of `(` and `)` into S.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      S
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  
  # Set max recursion limit
  sys.setrecursionlimit(1000000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  def li_input():
      return [int(_) for _ in input().split()]
  
  
  def gcd(n, m):
      if n % m == 0:
          return m
      else:
          return gcd(m, n % m)
  
  
  def gcd_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = gcd(v, L[i])
  
      return v
  
  
  def lcm(n, m):
      return (n * m) // gcd(n, m)
  
  
  def lcm_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = lcm(v, L[i])
  
      return v
  
  
  # Width First Search (+ Distance)
  def wfs_d(D, N, K):
      """
      D: 隣接行列(距離付き)
      N: ノード数
      K: 始点ノード
      """
  
      dfk = [-1] * (N + 1)
      dfk[K] = 0
  
      cps = [(K, 0)]
      r = [False] * (N + 1)
      r[K] = True
      while len(cps) != 0:
          n_cps = []
          for cp, cd in cps:
              for i, dfcp in enumerate(D[cp]):
                  if dfcp != -1 and not r[i]:
                      dfk[i] = cd + dfcp
                      n_cps.append((i, cd + dfcp))
                      r[i] = True
  
          cps = n_cps[:]
  
      return dfk
  
  
  # Depth First Search (+Distance)
  def dfs_d(v, pre, dist):
      """
      v:  現在のノード
      pre: １つ前のノード
      dist: 現在の距離
  
      以下は別途用意する
      D: 隣接リスト(行列ではない)
      D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
      """
  
      global D
      global D_dfs_d
  
      D_dfs_d[v] = dist
  
      for next_v, d in D[v]:
          if next_v != pre:
              dfs_d(next_v, v, dist + d)
  
      return
  
  
  def sigma(N):
      ans = 0
      for i in range(1, N + 1):
          ans += i
      return ans
  
  
  class Combination:
      def __init__(self, n, mod):
  
          g1 = [1, 1]
          g2 = [1, 1]
          inverse = [0, 1]
          for i in range(2, n + 1):
              g1.append((g1[-1] * i) % mod)
              inverse.append((-inverse[mod % i] * (mod // i)) % mod)
              g2.append((g2[-1] * inverse[-1]) % mod)
          self.MOD = mod
          self.N = n
          self.g1 = g1
          self.g2 = g2
          self.inverse = inverse
  
      def __call__(self, n, r):
          if r < 0 or r > n:
              return 0
          r = min(r, n - r)
          return self.g1[n] * self.g2[r] * self.g2[n - r] % self.MOD
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      N = int(input())
      S = input()
  
      open, close = 0, 0
  
      for i in range(N):
          if S[i] == "(":
              open += 1
          else:
              if open == 0:
                  close += 1
              else:
                  open -= 1
  
      print("(" * close + S + ")" * open)
  
  
  main()
  
problem_description: 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.
samples: [{"input": "3\n    ())", "output": "(())\n    \n\n* * *"}, {"input": "6\n    )))())", "output": "(((()))())\n    \n\n* * *"}, {"input": "8\n    ))))((((", "output": "(((())))(((())))"}]