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;
#define lli long long int
#define uli unsigned long long int
#define INF 999999999999999999
#define rep(i,m,n) for(lli i = m;i < n;i++)
#define rrep(i,m,n) for(lli i=m-1;i>=n;i--)
#define pb(n) push_back(n)
#define UE(N) N.erase(unique(N.begin(),N.end()),N.end());
#define Sort(n) sort(n.begin(), n.end())
#define Rev(n) reverse(n.begin(),n.end())
#define Out(S) cout << S << endl
#define NeOut(S) cout << S
#define HpOut(S) cout << setprecision(30) << S << endl
#define Vec(K,L,N,S) vector<L> K(N,S)
#define DV(K,L,N,M,S) vector<vector<L>> K(N,vector<L>(M,S))
#define TV(K,L,N,M,R,S) vector<vector<vector<L>>> K(N,vector<vector<L>>(M,vector<L>(R,S)))
#define pint pair<lli,lli>
#define paf(L,R) pair<L,R>
#define mod 1000000007
#define MAX 5100000
#define ALL(a)  a.begin(),a.end()
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
int main(){
  lli A,B,C,D,E,F,N,M,K,L,R,X,Y,Z,H,W,sum=0,num=0,flag=0;string S,T;
  cin >> N >> S;
  X=0;
  rep(i,0,N)if(S[i]=='(')sum++;else sum--;if(sum)X++;
  rep(i,0,X)S='('+S;
  rep(i,0,sum)S+=')';
  Out(S);
}