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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, j, n) for (ll i = j; i < n; i++)
#define all(x) (x).begin(),(x).end()
#define INF (1000000000)
#define MOD (1000000007)
#define MAX (100000)
#define pii pair<int, int>

/////////////////////////////////////////////////////////

class XY{
public:
  ll x, y;
  XY() {x = y = 0;}
  XY(ll u, ll v) {x = u; y = v;}
};

template<typename T1, typename T2>
void chmin(T1 &a, T2 b) {if(a > b) a = b;}
template<typename T1, typename T2>
void chmax(T1 &a, T2 b) {if(a < b) a = b;}

template<typename T1, typename T2>
T1 mypow(T1 a, T2 n){
  if(n == 0) return 1;
  if(n == 1) return a;
  if(n % 2) return a * mypow(a, n - 1);
  T1 tmp = mypow(a, n / 2);
  return tmp * tmp;
}

template<typename T>
int BS(vector<T> V, int left, int right, T key){
  int mid = (left + right) / 2;
  if(V[mid] <= key) left = mid;
  else right = mid;
  if(right - mid == 1) return left;
  else return BS(V, left, right, key);
}

/////////////////////////////////////////////////////////

void Main(){
  int N;
  string S;
  cin >> N >> S;
  int cnt1 = 0;
  int cnt2 = 0;
  int l = 0;
  int r = 0;
  rep(i, 0, N){
    if(S[i] == '('){
      l++;
    }else{
      r++;
      if(r > l) cnt1++;
    }
  }
  cnt2 = (l + cnt1) - r;
  rep(i, 0, cnt1) cout << '(';
  cout << S;
  rep(i, 0, cnt2) cout << ')';
  cout << endl;
}

/////////////////////////////////////////////////////////

int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << std::fixed << std::setprecision(15);

  Main();
}