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

(((())))(((())))
/*
* @Author Silviase(@silviasetitech)
* For ProCon
*/

import java.util.*;
import java.lang.*;
import java.math.*;

class Main{

    static int n;
    static int c;
    static int k;
    static int lef;
    static int rig;
    static String s;
    static String anss;
    static int[] times;
    static int m;
    static long ans;
    static int[] w;
    static int[] ww;
    static boolean[] visit;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        n = sc.nextInt();
        s = sc.next();

        anss = makekakko(s);

        System.out.println(anss);
        sc.close();
    }

    public static String makekakko(String s) {
        String ansss = s;
        int minn = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == ')' ){
                c--;
            }else{
                c++;
            }
            minn = Math.min(minn, c);
        }

        minn *= -1;

        for (int i = 0; i < minn; i++) {
            ansss = "("+ansss;
        }
        for (int i = 0; i < c+minn; i++) {
            ansss = ansss+")";
        }

        return ansss;
    }

    public static long gcd(long a, long b) {
        
        if (a < b) {
            long tmpp = a;
            a = b;
            b = tmpp;
        }
        if (b == 0) {
            return a;
        }else{
            return gcd(b, a%b);
        }
    }

    public static long lcm(long a, long b) {
        long gcd = gcd(a,b);
        return a/gcd*b;
    }

    public static void dfs(int placenow) {
        // if went all -> success! 
        // if not went all -> fail...
        
        /*
         dfs
         Go All Place that there is way to and not having been yet. 

         if island 1 is start point, dfs(1);
         if there is way to island 2 and island 3,
         - island 2 changes non-visited -> visited, and dfs(2).
         - island 3 changes non-visited -> visited, and dfs(3).
         

        */

        visit[placenow] = true;
        boolean success = true;
        for (int i = 0; i < n; i++) {
            if (visit[i] == false) { // not go all place
                success = false;
                break;
            }
        }
        if (success) {
            ans++;
            visit[placenow] = false;
            return;
        }

        for (int i = 0; i < m; i++) {
            if (w[i] == placenow && visit[ww[i]] == false ) {
                dfs(ww[i]);
            }else if(ww[i] == placenow && visit[w[i]] == false){
                dfs(w[i]);
            }else{
                continue;
            }
        }
        visit[placenow] = false;
        return;
    }

}

