A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You are given a regular bracket sequence s and an integer number k. Your task is to find a regular bracket sequence of length exactly k such that it is also a subsequence of s.

It is guaranteed that such sequence always exists.

Input

The first line contains two integers n and k (2 ≤ k ≤ n ≤ 2 ⋅ 10^5, both n and k are even) — the length of s and the length of the sequence you are asked to find.

The second line is a string s — regular bracket sequence of length n.

Output

Print a single string — a regular bracket sequence of length exactly k such that it is also a subsequence of s.

It is guaranteed that such sequence always exists.

Examples

Input

6 4
()(())


Output

()()


Input

8 8
(()(()))


Output

(()(()))
import java.util.Scanner;

public class C1023 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		int n = in.nextInt(), len = in.nextInt(), i, toRemove = 0, left = 0, resultIndex = 0;
		char[] s = in.next().toCharArray();
		char[] result = new char[len];
		
		toRemove = (n - len) / 2;
		for(i = 0; i < n; ++i) {
			if(s[i] == '(' && toRemove > 0) {
				left++;
				toRemove--;
				continue;
			}
			if(s[i] == '(' && toRemove == 0) {
				result[resultIndex++] = '(';
				continue;
			}
			if(s[i] == ')' && left > 0) {
				left--;
				continue;
			}
			if(s[i] == ')' && left == 0)
				result[resultIndex++] = ')';
		}
		
		System.out.println(result);
	}
}
