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.

Examples

Input

3
())


Output

(())


Input

6
)))())


Output

(((()))())


Input

8
))))((((


Output

(((())))(((())))
import java.util.*;
import java.util.Map.Entry;
 
 
 
public class Main{
	
	
	
	public static Set<Long> set=new HashSet<Long>();
	public static long T;
	public static long ans;
	

 
	
	//最大公約数・最小公倍数（セットで使う）
	public static long gcm(long a, long b) {
		long result = a;
		long k = 0;
		long n = b;
		do {
			k = result % n;
			result = n;
			n = k;
		} while(k != 0);
		return result;
	}
	
	public static long lcm(long a, long b) {
		long g;
		g = gcm(a, b);
		return a/g*b;
	}
	
	static long gcd(long a, long b) {
		if (b == 0) {
			return a;			
		}
		return gcd(b, a % b);
	}
	
	//等差数列
	static long tousasum(int n) {
		return (long)n*((long)n+1)/2;
	}
 
 
 
	//素数判定
	static boolean isPrime (int n) {
		if (n==2) return true;
		if (n<2 || n%2==0) return false;
		double d = Math.sqrt(n);
		for (int i=3; i<=d; i+=2) if(n%i==0){return false;}
		return true;
	}
 
 
	//倍数判定（10進数以外のときに有用）
	static boolean isMultiple (String s, int base, int m) {
		int temp = 0;
		for (int i=0; i<s.length(); i++) {
			temp = (temp*base+Character.getNumericValue(s.charAt(i)))%m;
		}
		if (temp==0) {return true;}
		return false;
	}
 
 
	//階乗
	static long factorial (int i) {
		if (i==1) {return 1;}
		else {return i*factorial(i-1);}
	}
 
 
	//進数変換
	static String toNbase (String sm, int m, int n) {
		return Long.toString(Long.parseLong(sm,m),n);
	}
	
	public static final long gcd_n(long[] param) {
	    final int len = param.length;
	    long g = gcd(param[0], param[1]);    //gcd(a, b) は以前作ったもの
	    for (int i = 1; i < len - 1; i++) {
	        g = gcd(g, param[i + 1]);       //gcd(a, b) は以前作ったもの
	    }
	    return g;
	}
	
	public static long lcm_n(long[] numbers) {
		long l;
		l = numbers[0];
		for (int i = 1; i < numbers.length; i++) {
			l = lcm(l, numbers[i]);
		}
		return l;
	}
	
    public static boolean tree(long N,long r) {
    	ans = r;
    	if(N==T) {
    		return true;
    	}else if(2*N==T||2*N+1==T) {
    		ans = r+1;
    		return true;
    	}else if(2*N>T) {
    		return false;
    	}
    	return tree(2*N,r+1)||tree(2*N+1,r+1);
    	
    	
    }
	
	
 public static void main(String[] args) {
	 Scanner sc=new Scanner(System.in);
	 int N=sc.nextInt();
	 String s=sc.next();
	 char[] c=s.toCharArray();
	 
	 int left = 0;
	 int right = 0;
	 
	 String ans=s;
	 
	 for(int i=0;i<N;i++) {
		 if(c[i]=='(') {
			 right++;
			 
		 }else {
			 if(right==0) {
				 left++;
			 }else {
				 right--;
			 }
		 }
		
			 
			
	 }
	 
	 for(int j=0;j<left;j++) {
		 ans="("+ans;
	 }
	 
	 for(int j=0;j<right;j++) {
		 ans+=")";
	 }
	 
	 System.out.println(ans);
	 
    
}
}

