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>
using namespace std;
const long long INF = 1 << 31 - 1, MAX = 2e5 + 100;
long long a, b, x, y, n, k, m, cnt, ans, mmax, tmp, itor;
long long stk[MAX];
long long use[MAX];
string s;
signed main() {
  cin >> n >> k >> s;
  long long dir = n - k;
  cnt = 1;
  for (long long i = 0; i < n; ++i) {
    if (s[i] == '(') {
      use[i] = cnt;
      stk[itor] = i;
      use[i] = cnt;
      cnt++, itor++;
    } else {
      if (itor >= 0) {
        use[i] = use[stk[itor - 1]];
        itor--;
      } else {
        use[i] = -1;
        dir--;
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    if (use[i] > 0 && use[i] <= k / 2) cout << s[i];
  }
  cout << endl;
  return 0;
}
