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 decimal,math
from collections import *
from fractions import gcd
from bisect import bisect_right,bisect_left
import sys

decimal.getcontext().prec = 15

def primeFactors(n): 
	arr =set([])
	while n % 2 == 0: 
		arr.add(2) 
		n = n / 2
	for i in xrange(3,int(math.sqrt(n))+1,2): 
		while n % i== 0: 
			arr.add(i) 
			n = n / i 
	if n > 2: 
		arr.add(n)
		# print n
	return arr	 
def z_advanced(s):
	Z = [0] * len(s)
	Z[0] = len(s) 
	rt = 0
	lt = 0
	for k in xrange(1, len(s)):
		if k > rt:
			n = 0
			while n + k < len(s) and s[n] == s[n+k]:
				n += 1
			Z[k] = n
			if n > 0:
				lt = k
				rt = k+n-1
		else:
			p = k - lt  # Pair index.
			right_part_len = rt - k + 1 
			if Z[p] < right_part_len:
				Z[k] = Z[p]
			else:
				i = rt + 1
				while i < len(s) and s[i] == s[i - k]:
					i += 1
				Z[k] = i - k
 
				lt = k
				rt = i - 1
	return max(Z)
def invmod(i,mod):
	return pow(i,mod-2,mod)	
def fact(mod):
	fac= [1,1]
	ifac = [1,1]
	for i in xxrange(2,(3*10**5+ 2)):
		fac.append((fac[-1]*i)%mod)
		# ifac.append((ifac[-1]*invmod(i,mod))%mod)
	return fac,ifac	
def SieveOfEratosthenes(n): 
	prime = [True for i in xrange(n+1)] 
	p = 2
	while (p * p <= n): 
		if (prime[p] == True): 
			
			for i in xrange(p * p, n+1, p): 
				prime[i] = False
		p += 1
	arr=[]
	for p in xrange(2, n): 
		if prime[p]: 
			arr.append(p)
	return arr
def maxSubArraySum(a):
	size=len(a)       
	max_so_far =a[0] 
	curr_max = a[0] 
	for i in xrange(1,size): 
		curr_max = max(a[i], curr_max + a[i]) 
		max_so_far = max(max_so_far,curr_max) 
	return max_so_far 	
def fi():
	return int(sys.stdin.readline())
 
def fi2():
	return map(int, sys.stdin.readline().split())
 
def fi3():
	return sys.stdin.readline()
 
def fo(*args):
	for s in args:
		sys.stdout.write(str(s)+' ')
	sys.stdout.write('\n')
n,k = fi2()
s = raw_input()
count = 0
for i in range(len(s)):
	if s[i] == '(':
		count= count + 1
	if count == k/2:
		break
ans = s[:i+1] + ')'*(k-i-1)		
print ans