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 BraquetSequence {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int n = scan.nextInt() , m = scan.nextInt();
		
		char[] arr = scan.next().toCharArray();
		
		StringBuilder sb = new StringBuilder();
		
		int op = 0 ;
		int cl = 0 ;
		
		sb.append(arr[0]);
		op++;
		
		for (int i = 1; i < arr.length; i++) {
			char c = arr[i];

			if(op == m/2) {

				while(cl++<m/2)sb.append(')');
				break;
				
			}
			
			if(op > cl)
			{
					sb.append(c);
					if(c == ')')cl++;		
					else op++;
					
			}
	
			
			if(op == cl)
			{
				if(c == '(') {
					op++;
					sb.append(c);
				}
			}
		}
		System.out.println(sb);
	}
}
