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.ArrayList;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int N = sc.nextInt();
		sc.nextLine();

		String given = sc.nextLine();
		
		// 与えられた括弧の列を、ArrayListに格納する
		ArrayList<Character> problem = new ArrayList<Character>();
		for (int i = 0; i < given.length(); i++) {
			problem.add(given.charAt(i));
		}
		
		// 解答用のコレクションを作成
		// 本当はうまく problem をコピーしたかったが、できなかったので（referenceになったので）、再度作成している
		ArrayList<Character> answer = new ArrayList<Character>();
		for (int i = 0; i < given.length(); i++) {
			answer.add(given.charAt(i));
		}
		
		// 与えられた文字列から、正しい括弧列を削除する。
		// 文字列を一文字ずつ走査し、"()" が見つかったら、これを削除し、最初から走査する。
		// 文字列がなくなるか、最後の文字に到達したら終了。
		int index = 0;
		while (true) {
			if (problem.size() == 0) {
				break;
			}
			
			if (index == problem.size()) {
				break;
			}
			
			if (index == 0) {
				index++;
				continue;
			}
				
			if (problem.get(index - 1) == '(' && problem.get(index) == ')') {
				problem.remove(index);
				problem.remove(index - 1);
				// インデックスを最初に戻す
				index = 0;
				continue;
			}
			index++;
		}

		// "(" と ")" の数を数える。
		int left = 0;
		int right = 0;
		for (int i = 0; i < problem.size(); i++) {			
			if(problem.get(i) == '(') {
				left++;
			} else {
				right++;
			}
		}
		// debug
		// System.out.println("left: " + left + " right: " + right);
		// System.out.println(answer.size());
		
		
		// ")" の数だけ配列の先頭に "(" を加え、"(" の数だけ配列の末尾に ")" を加える。
		for (int i = 0; i < right; i++) {
			answer.add(0, '(');
		}
		for (int i = 0; i < left; i++) {
			answer.add(')');
		}
		
		for (int i = 0; i < answer.size(); i++) {
			System.out.print(answer.get(i));
		}
	}	
}
