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

(()(()))
#include <bits/stdc++.h>
#pragma GCC optimize(O3)
using namespace std;
long long c = 0, d, e, a, b, res = 1e18, mas[200001], mas2[200000], f, z, res2,
          mas3[1000], array57[10][100000], mas4[100], r1, r2;
long long base = 57, p = 1e9 + 7, ans = 10000, sum, sum2, uk1, uk2, heh, heh2;
char ch;
string s, s2, s3, s4;
set<pair<long long, long long>> mset;
vector<pair<long long, long long>> vec;
deque<long long> ord, answer[5000], vec2;
map<pair<long long, long long>, long long> map57;
multiset<long long> mset57;
long long map1[5000], map2[5000][101];
long long get_sum(vector<long long> a) {
  long long answer2 = 0, r = 1;
  for (int i = 0; i < a.size(); i++) {
    answer2 += r * a[i];
    r *= 2;
  }
  return answer2;
}
signed main() {
  srand(time(0));
  cin >> a >> b;
  cin >> s;
  for (int i = 0; i < a; i++) {
    if (s[i] == '(') {
      mas[c] = d;
      mas2[i] = d;
      c++;
      d++;
    } else {
      c--;
      mas2[i] = mas[c];
    }
  }
  e = a - b;
  e /= 2;
  for (int i = 0; i < a; i++) {
    if (mas2[i] >= e) {
      cout << s[i];
    }
  }
  cin >> a;
  return 0;
}
