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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.nextLine());
        String s = sc.nextLine();

        String ans = solve(n, s);
        System.out.println(ans);
    }

    static String solve(int n, String s) {
        int cnt = countPare(s);
        if (cnt == 0 && s.charAt(0) == '(') {
            // 括弧の誤差無しで、1文字目が開き括弧である
            return s;
        }

        String ans = "";
        if (cnt < 0) {
            ans = insert(s, "(", Math.abs(cnt));
        } else if (cnt > 0) {
            ans = insert(s, ")", Math.abs(cnt));
        } else {
            ans = insert(s, ")", s.length() / 2);
            ans = insert(ans, "(", s.length() / 2);
        }

        return ans;
    }

    static String insert(String origin, String kakko, int cnt) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < cnt; i++) {
            sb.append(kakko);
        }
        if (kakko.equals(")")) {
            // 末尾に挿入
            return origin.concat(sb.toString());
        } else {
            // 先頭に挿入
            return sb.toString().concat(origin);
        }
    }

    static int countPare(String s) {
        char[] c = s.toCharArray();
        int open = 0;
        int close = 0;
        for (int i = 0; i < c.length; i++) {
            if (c[i] == '(') {
                open++;
            } else {
                close--;
            }
        }
        return open + close;
    }
}