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

(()(()))
def delete(s,pos):
        return str(s[:pos]+s[(pos+1):])
n,m=[int(i) for i in input().split()]
a=str(input())
if n==m:
        print(a)
else:
        m=m//2
        t=a.count('()')
        z=a.count('(')
        if m<=t:
                g='()'*m
                print(g)
        elif (z-t)>=m:
                #print(z,t,m)
                g='('*m + ')'*m
                print(g)
        else:
                while(n!=2*m):
                        x=a.find('(')
                        y=a.find(')')
                        #print(x,y)
                        a=delete(a,x)
                        a=delete(a,y-1)
                        #print(a)
                        n=n-2
                print(a)
                
                
