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: Accepted)

//#include <bits/stdc++.h>
//#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>


#define FOR(i, a, b) for( int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for( int i=(a);i>=(b);i--)
#define LFOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define LRFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define INF 1000000000 //2000000000
#define LINF 1000000000000000000 //9000000000000000000
#define PI 3.14159265358979
#define MAXI 7500000

using namespace std;
typedef long long int ll;
typedef pair< long long int, long long  int> P;

int dy[5] = { 0,0,1,-1,0 };
int dx[5] = { 1,-1,0,0,0 };

int main(void){
    int n;
    string s;
    int r = 0;
    int l = 0;
    int nowr = 0;
    int nowl = 0;

    cin >> n;
    cin >> s;

    FOR(i,0,s.size()-1){
        if(s[i]=='('){
            l++;
        }else{
            r++;
        }
    }
    FOR(i,0,s.size()-1){
        if(s[i]=='('){
            nowl++;
        }else{
            nowr++;
            if(nowl>0){
                nowl--;
                nowr--;
            }
        }
    }
    FOR(i,1,nowr){
        cout << "(";
    }
    cout << s;
    FOR(i,1,nowl){
        cout << ")";
    }
    cout << endl;
}